9

I. Third Powers

As background, the complete solution to $x_1^3+x_2^3 = x_3^3+x_4^3$ was given by Euler and Binet using deg-$4$ polynomials. However, Elkies also found a complete solution using deg-$3$ polynomials in $3$ variables. Define.

$$f(a,b,c) = 9a^3 + 9a^2b + 3a^2c + 3a b^2 - 6a b c + 3a c^2 + 3b^3 + 3b^2c + b c^2 + c^3$$

then,

$$f(a,b,c)^3+f(-a,b,-c)^3=f(a,b,-c)^3+f(-a,b,c)^3$$


II. Fourth Powers

In contrast, no complete polynomial solution to $x_1^4+x_2^4 = x_3^4+x_4^4$ is known (or maybe even possible). However, analogous to Elkies' solution, there are at least five of form,

$$f(a, b)^4 + f(b, -a)^4 = f(a, -b)^4 + f(b, a)^4$$

where $f(a,b)$ is a polynomial of degree $k = 6n+1 = 7,13,19,25,31$ given in the addendum below. Whether there are more remains to be seen. (Deg-$25$ was found by yours truly.)


III. Elliptic curve

Using Euler's approach, let,

$$(p+q)^4+(r-s)^4=(p-q)^4+(r+s)^4$$

and define $p = (a^3 - b),\, q = a y,\, r = b (a^3 - b),\, s = y.\,$ Then the equation above transforms to the simple form,

$$(a^3 - b) (b^3 - a) = y^2$$

This is birationally equivalent to an elliptic curve. (Update: This curve has been analyzed by Deyi Chen in this MO answer.) Known rational points of small height are,

$$a=n,\quad b =n\,\frac{(4 + n^2 + 10n^4 + n^6)}{(1 + 10n^2 + n^4 + 4n^6)}\quad$$

$$a=n^3,\quad b =n\,\frac{(1 - 2n^2 + n^4 + n^6)}{(1 + n^2 - 2n^4 + n^6)}\quad$$

$$\quad a=n,\quad b =n\,\frac{(9 - 44n^2 + 190n^4 + 100n^6 + n^8)}{(1 + 100n^2 + 190n^4 - 44n^6 + 9n^8)}$$

$$\quad a=n^3,\quad b =n\,\frac{(1 - 4n^2 + 10n^6 + 8n^{10} + n^{12})}{(1 + 8n^2 + 10n^6 - 4n^{10} + n^{12})}$$

Note the symmetry of $b$'s numerator and denominator. The first two points $b$ yield the same deg-$7$ solution (after removing common factors) while the last two $b$ yield the deg-$13$ and deg-$19$ solutions. For the deg-$25$, we have $a=n,$ and,

$\qquad\qquad b =\frac{n(16 - 543n^2 + 4632n^4 + 15100n^6 + 10632n^8 + 22758n^{10} + 6568n^{12} + 5820n^{14} + 552n^{16} + n^{18})}{(1 + 552n^2 + 5820n^4 + 6568n^6 + 22758n^8 + 10632n^{10} + 15100n^{12} + 4632n^{14} - 543n^{16} + 16n^{18})}$

found by yours truly (though I may have missed a point of smaller height). The deg-$31$ solution has a rational point $b$ that is of higher height.


IV. Question

Q: Given some fixed $a$ (like $a=n$) and the elliptic curve $(a^3 - b)(b^3 - a) = y^2,$ can you find a rational point $b = P(n)/Q(n)$ that is of height less than deg-$24$?

Note: Points that yield an identity of form $f(a, b)^4 + f(b, -a)^4 = f(a, -b)^4 + f(b, a)^4$ are preferred. An example of a point that does not do so was found by Lander (and which leads to a second deg-$19$ identity).


Addendum:

Below are the five known polynomials $f(a,b)$ of deg $k = 6n+1 = 7,13,19,25,31.$ (For ease of copy-paste, they are purposely not in Latex format for those who want to test them.)

The deg-$7$, or versions thereof, have been independently found by many including Euler, Gerardin, Swinnerton-Dyer (at the tender age of 16), and others.

Deg 7

f(a, b) = a^7 + a^5 b^2 - 2 a^3 b^4 - 3 a^2 b^5 + a b^6 

Deg 13

f(a, b) = a^13 + a^12 b - a^11 b^2 + 5 a^10 b^3 - 6 a^9 b^4 - 12 a^8 b^5 + 4 a^7 b^6 +  7 a^6 b^7 + 3 a^5 b^8 - 3 a^4 b^9 - 4 a^3 b^10 + 2 a^2 b^11 + a b^12 + b^13

Deg 19

f(a, b) = a^19 + 6 a^17 b^2 - 18 a^15 b^4 + 6 a^14 b^5 - 5 a^13 b^6 + 12 a^12 b^7 - 12 a^11 b^8 + 36 a^10 b^9 - 24 a^9 b^10 - 12 a^8 b^11 + 19 a^7 b^12 + 36 a^6 b^13 + 6 a^5 b^14 + 12 a^4 b^15 - 6 a^3 b^16 + 6 a^2 b^17 + a b^18

Deg 25

f(a, b) = a^25 + 7 a^23 b^2 - 2 a^21 b^4 - 3 a^20 b^5 - 25 a^19 b^6 - 63 a^18 b^7 +  43 a^17 b^8 + 36 a^16 b^9 - 134 a^15 b^10 + 213 a^14 b^11 + 179 a^13 b^12 - 333 a^12 b^13 - 128 a^11 b^14 + 207 a^10 b^15 + 136 a^9 b^16 - 57 a^8 b^17 - 97 a^7 b^18 + 9 a^6 b^19 + 4 a^5 b^20 - 9 a^4 b^21 + 10 a^3 b^22 - 3 a^2 b^23 + a b^24

Deg 31

f(a, b) = a^31 - a^30 b + 11 a^29 b^2 + a^28 b^3 + 42 a^27 b^4 + 24 a^26 b^5 - 19 a^25 b^6 - 32 a^24 b^7 - 154 a^23 b^8 - 254 a^22 b^9 + 266 a^21 b^10 +  718 a^20 b^11 + 126 a^19 b^12 - 303 a^18 b^13 - 478 a^17 b^14 - 830 a^16 b^15 + 770 a^15 b^16 + 916 a^14 b^17 - 738 a^13 b^18 +  21 a^12 b^19 + 350 a^11 b^20 - 434 a^10 b^21 + 50 a^9 b^22 + 142 a^8 b^23 - 91 a^7 b^24 + 76 a^6 b^25 + 15 a^5 b^26 - 3 a^4 b^27 + 8 a^3 b^28 - 8 a^2 b^29 + a b^30 - b^31

P.S. A similar question was asked by emacs in this post.

  • Do we know of just one numerical case to get an idea? – user25406 Oct 03 '23 at 10:25
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    @user25406: No. Euler, by sheer algebraic manipulation, found the deg-$7$ and deg-$13$ solutions. But he's Euler, after all. – Tito Piezas III Oct 03 '23 at 10:52
  • Would that mean that solution(s) will be found (if they exist) only with numbers with a huge number of digits? – user25406 Oct 03 '23 at 10:56
  • @emacsdrivesmenuts Yes, I mentioned your post at the end of mine. :) – Tito Piezas III Oct 03 '23 at 12:15
  • @Tito, I wasn't able to find a solution with a low degree, but I found a parametric solution for $k = 97.$ – Tomita Oct 04 '23 at 07:56
  • @Tomita Is it of form $f(a, b)^4 + f(b, -a)^4 = f(a, -b)^4 + f(b, a)^4$? – Tito Piezas III Oct 04 '23 at 08:17
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    @Tito, Yes, I guess that's probably true. The first few terms are $A=a^{97}+13b^2a^{95}+25b^4a^{93}+9b^5a^{92}-318b^6a^{91}+621b^7a^{90}-1726b^8a^{89}\cdots$ $B=-b^{97}-13a^2b^{95}-25a^4b^{93}+9a^5b^{92}+318a^6b^{91}+621a^7b^{90}+1726a^8b^{89}\cdots$ $C=a^{97}+13b^2a^{95}+25b^4a^{93}-9b^5a^{92}-318b^6a^{91}-621b^7a^{90}-1726b^8a^{89}\cdots$ $D=b^{97}+13a^2b^{95}+25a^4b^{93}+9a^5b^{92}-318a^6b^{91}+621a^7b^{90}-1726a^8b^{89}\cdots$ – Tomita Oct 04 '23 at 09:42
  • @Tomita Nice! Did you use the elliptic curve above? If so, what was the degree of the polynomial in $b$? The numerator and denominator were symmetric like the others, weren’t they? – Tito Piezas III Oct 04 '23 at 09:56
  • @Tito, Yes, I used previous result. A^4+B^4=C^4+D^4 The result was derived from $m = \frac{a^3+b^3}{-b^3+a^3}$ using group law. – Tomita Oct 04 '23 at 10:28
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    @Tito, We use a known point $P(m, v) = (\frac{a^3+b^3}{-b^3+a^3}, \frac{24a^3b^3(a+b)}{(a^2+ab+b^2)^2(-b+a)}).$ The result was derived from $3P.$ The elliptic curve is $v^2 = (-72b^2a^2+64a^3b-28b^4+64b^3a-28a^4)m^4+(64a^4-64a^3b-64b^4+64b^3a)m^3+(-72a^4+144b^2a^2-72b^4)m^2+(64a^3b+64a^4-64b^3a-64b^4)m-28b^4-64a^3b-28a^4-64b^3a-72b^2a^2.$ – Tomita Oct 04 '23 at 10:51
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    Hello, dear friend, this is a wonderful question, the framework, and the motivation for the question present an interesting with many explicit solutions, upvoted! I am trying now to understand the exactly question. So far i understand it as follows. Consider the curve in $(u,v)$ with parameter $A$ / over the function field $\Bbb Q(A)$ given by:$$v^2=(u^3-A)(A^3-u)\ .$$The Question asks now - as i understand - for a $\Bbb Q(A)$-rational point $(u_0(A),v_0(A)$ (with $b$ from the question being a/the rational function $u_0(A)$). If so, specializations of $A$ would lead to curves with points. – dan_fulea Oct 04 '23 at 11:58
  • But many specializations have rank zero and no torsion points over $\Bbb Q$. To reproduce this, substitute in the second factor on the R.H.S. $$A^3-u=\frac 1x$$, so $u=A^3-\frac 1x$. Let $v=y/x^2$ to have a cancelling denominator. Set $X=A^3x-1$. Then we obtain the curve $$y^2=X^3-A\ .$$For many values of $A$ (that avoid possible poles in any fixed rational functions $u_0,v_0$ in $\Bbb Q(A)$) we have rank zero and no torsion points... – dan_fulea Oct 04 '23 at 12:03
  • @dan_fulea Well, we can have two parameters $(u,,A)$. We can set $u = n$ and $A = P_1(n)$. Or set $u = n^3$ and $A = P_2(n)$. For each case, I have given initial polynomial solutions $P(n)$ which should lead to further ones. However, we are hoping for $P(n)$ with "smallish" degree, since I'm sure I skipped some. – Tito Piezas III Oct 04 '23 at 12:09
  • Yes, this is in the spirit of the examples, where both $u,A$ may be elements of some function field $\Bbb Q(n)$. But the question is explicitly: "Given some >>>fixed $a$<<< and the elliptic curve..." - so i am asking before starting a laborious search. The fixed $a$ was in the sample solutions not "generally" fixed. As i understand it now, we are searching for polynomials $v,a,b$ in $\Bbb Q[n]$ so that $v^2=(a^3-b)(b^3-a)$, with $a,b$ (or at least one of them) having degree $\le 24$. Is it so? – dan_fulea Oct 04 '23 at 12:21
  • @dan_fulea Correct. But there is the added constraint that these "special" polynomial solutions must lead to an identity of form $f(a, b)^4 + f(b, -a)^4 = f(a, -b)^4 + f(b, a)^4$ (as not all solutions will qualify and which makes the problem challenging). Let us discuss the simplest case $(a^3-b)(b^3-a) = y^2$ where $a=n$. I found three polynomial solutions $b=P(n)$ where the numerator has degree $7,9,19$. Did I find all with degree $<24$, or did I miss some? – Tito Piezas III Oct 04 '23 at 12:42
  • @dan_fulea I just found the identity for degree $k=37$. Kindly see my partial answer below. – Tito Piezas III Oct 04 '23 at 14:35
  • Will rewriting the equation as $(a^2)^2 + (b^2)^2 = (c^2)^2 + (d^2)^2$ and convert it to $(a^2)^2 - (c^2)^2 = (d^2)^2 - (b^2)^2$ then using $(a^2 - c^2)(a^2 +c^2) = (d^2-b^2)(d^2 +b^2)$ help in any way? – user25406 Oct 08 '23 at 11:11

5 Answers5

3

(This is more a long comment.)

The question seeks to find special identities of form,

$$f(\alpha, \beta)^4 + f(\beta, -\alpha)^4 = f(\alpha, -\beta)^4 + f(\beta, \alpha)^4 \tag{1}$$

Known ones have degree $k = 6n+1 = 7,13,19,25,31.$ However, additional ones have been found today for $k=37$ (Piezas) and $k = 97$ (Tomita). One way is by solving the rather simple elliptic curve,

$$(a^3 - b) (b^3 - a) = y^2$$


I. First family ($a = n$)

We now have five six "smallish" solutions,

$b_1 =\frac{n\,(\color{red}{1} + n)}{(1 + n)}$

$b_2 =\frac{n\,(\color{red}{4} + n^2 + 10n^4 + n^6)}{(1 + 10n^2 + n^4 + 4n^6)}$

$b_3 =\frac{n\,(\color{red}{9} - 44n^2 + 190n^4 + 100n^6 + n^8)}{(1 + 100n^2 + 190n^4 - 44n^6 + 9n^8)}$

$b_4 =\frac{n\,(\color{red}{16} - 543n^2 + 4632n^4 + 15100n^6 + 10632n^8 + 22758n^{10} + 6568n^{12} + 5820n^{14} + 552n^{16} + n^{18})}{(1 + 552n^2 + 5820n^4 + 6568n^6 + 22758n^8 + 10632n^{10} + 15100n^{12} + 4632n^{14} - 543n^{16} + 16n^{18})}$

$b_5 =\frac{n\,(\color{red}{25} -3524n^2 + 113482n^4 + 979388n^6 + 1486687n^8 + 2379064n^{10} + 5807660n^{12} + 3492760n^{14} + 2404327n^{16} + 45836n^{18} + 69418n^{20} +2092n^{22} + n^{24})\quad}{(1 + 2092n^2 + 69418n^4 + 45836n^6 + 2404327n^8 + 3492760n^{10} + 5807660n^{12} + 2379064n^{14} + 1486687n^{16} + 979388n^{18} + 113482n^{20} - 3524n^{22} + 25n^{24})}$

yielding identities of degree $1,7,13,25,37.$

Note: Tomita just found a sixth solution $b_6$ yielding the identity for $k=55$. The numerator of $b_6$ has the linear term $\color{red}{36}$, fitting the pattern above. Also, in the comments section, Tomita found $k=97$ using a different approach described here.


II. Second family ($a=n^3$)

We now have two four "smallish" solutions,

$b_1 =n\,\frac{(1 - 2n^2 + n^4 + n^6)}{(1 + n^2 - 2n^4 + n^6)}$

$b_2 =n\,\frac{(1 - 4n^2 + 10n^6 + 8n^{10} + n^{12})}{(1 + 8n^2 + 10n^6 - 4n^{10} + n^{12})}$

$b_3 = \frac{n(n^{30}+17n^{28}-18n^{26}+101n^{24}-172n^{22}+80n^{20}+282n^{18}-82n^{16}-244n^{14}+282n^{12}-28n^{10}-64n^8+101n^6+9n^4-10n^2+1)}{(n^{30}-10n^{28}+9n^{26}+101n^{24}-64n^{22}-28n^{20}+282n^{18}-244n^{16}-82n^{14}+282n^{12}+80n^{10}-172n^8+101n^6-18n^4+17n^2+1)}$

$b_4 = \frac{n(n^{48} + 32n^{46} + 552n^{42} - 1088n^{40} - 16n^{38} + 5820n^{36} + 8160n^{34} + 544n^{32} + 6552n^{30} + 18560n^{28} - 4080n^{26} + 23302n^{24} + 8160n^{22} - 9280n^{20} + 6552n^{18} - 1088n^{16} - 4080n^{14} + 5820n^{12} + 32n^{10} + 544n^8 + 552n^6 - 16n^2 + 1)}{n^{48} - 16n^{46} + 552n^{42} + 544n^{40} + 32n^{38} + 5820n^{36} - 4080n^{34} - 1088n^{32} + 6552n^{30} - 9280n^{28} + 8160n^{26} + 23302n^{24} - 4080n^{22} + 18560n^{20} + 6552n^{18} + 544n^{16} + 8160n^{14} + 5820n^{12} - 16n^{10} - 1088n^8 + 552n^6 + 32n^2 + 1}$

with the 3rd by Seiji Tomita and the 4th by Deyi Chen in this MO post. These yields identities of degree $7,19,43$.

P.S. Both families are peculiar in that the coefficients of the numerator and denominator are palindromes of each other.

  • Starting from a solution of the shape $(a(n),b(n))$ how do you compute the corresponding $f=f(a,b)$ - if any? – dan_fulea Oct 05 '23 at 22:05
  • @dan_fulea We use Euler's method, $$(p+q)^4+(r-s)^4 = (p-q)^4+(r+s)^4\tag1$$ where $(p,q,r,s)$ are defined by $(a,b,y)$ as described in the post section "III. Elliptic curve". Then, after removing common factors, we observe $(1)$ if it has the form $$f(\alpha, \beta)^4 + f(\beta, -\alpha)^4 = f(\alpha, -\beta)^4 + f(\beta, \alpha)^4$$ It seems to be the case if $b$'s numerator and denominator are palindromic wrt to each other. P.S. To prevent confusion, I realize it is best to use the expression $f(\alpha, \beta)$. I only used $f(a,b)$ following http://euler.free.fr/identities.htm. – Tito Piezas III Oct 06 '23 at 01:45
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    The sum of coefficients of the second family is also a power of 2 – Sidharth Ghoshal Oct 06 '23 at 04:24
  • @dan_fulea For the first family, it seems we've found the first NINE rational points except $b_8$. Can you kindly check this MO post and if you can shed some light on the matter? – Tito Piezas III Oct 06 '23 at 06:45
  • I still cannot catch the point (so that i can check by code the above cases and the ones in the OP and potentially new ones). Please help. So far i understand the following. In the sample situation with $a=a(n)=n$ and $b=b(n)=n g(1,n)/g(n,1)$, $g$ homogeneous with $g(n,1)=1+10n^2+n^4+4n^6$, we compute $y=y(n)=\pm\sqrt{(a^3-b)(b^3-a)}$, then plug in into $p,q,r,s$ defined respectively as $a^3-b$, $ay$, $b(a^3-b)$, $y$, some "common factors" are eliminated, and i get $p=8n^6 + 2n^4 + 20n^2 + 2$, $q=n^7 - 17n^5 - 17n^3 + n$, $r=2n^7 + 20n^5 + 2n^3 + 8n$, $s=n^6 - 17n^4 - 17n^2 + 1$. What is $f$? – dan_fulea Oct 06 '23 at 12:32
  • @dan_fulea You are near the point. Given your $(p,q)$, then $$p-q = 2 - n + 20n^2 + 17n^3 + 2n^4 + 17n^5 + 8n^6 - n^7$$ Then let $n=\frac{a+b}{a-b}$ to simplify and reduce the size of the coefficients, the above transforms to, $$f(a,b) =a^7 + a^5 b^2 - 2 a^3 b^4 - 3 a^2 b^5 + a b^6$$ and, $$f(a, b)^4 + f(b, -a)^4 = f(a, -b)^4 + f(b, a)^4$$ since the LHS and RHS can be switched. hope this explains the procedure. – Tito Piezas III Oct 06 '23 at 13:05
  • I am trying now to arrange the things so that:$$\begin{aligned}p+q &=f(n,1)&&\text{ with coeffs }[1,8,-17,2,-17,20,1,2]\r-s &=f(1,-n)&&\text{ with coeffs }[1,8,-17,2,-17,20,1,2]^{\text{reversed}}-\p-q &=f(-n,1)&&\text{ with coeffs }[1,8,-17,2,-17,20,1,2]-\r+s &=f(1,n)&&\text{ with coeffs }[1,8,-17,2,-17,20,1,2]^{\text{reversed}}\ .\end{aligned}$$The sub index minus is a $(-1)$ twist. So the produced polynomial is an other one... It has a zero in one. I tried to substitute $n\to n\pm 1$ or so... – dan_fulea Oct 06 '23 at 13:09
  • @dan_fulea Do the transformation $n = \frac{a+b}{a-b}$ (or whichever variable pairs you prefer) to simplify and reduce the size of the coefficients. It's a nice trick I stumbled upon. Everything ok now? – Tito Piezas III Oct 06 '23 at 13:12
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    OK, i see now how the mentioned polynomials occur, the missing information was that i have to start with the expression for $(p-q)$ as a function of $n$, and apply the Möbius transformation which replaces $n$ by $(n+1)/(n-1)$, then clear the denominators which are powers of $(n-1)$... i can now reproduce... – dan_fulea Oct 06 '23 at 13:17
  • @dan_fulea I asked Tomita to add to his answer the first term of the deg-$97$ identity, and I managed to derive the missing $b_8$ from it. (I suspected a pattern.) So I guess what remains for the MO question is to consider the general case. – Tito Piezas III Oct 06 '23 at 13:49
3

This is a long comment.

I feel like there are infinite rational solutions to $y^2 = (a^3-b)(b^3-a)$ using group law.


I. New results for first family.

I found an identity for $k=55$ using a solution of degree $7.$

$a=n$,

$ b_6 = \frac{(n^{38}+6234n^{36}+569433n^{34}-1574764n^{32}+165024372n^{30}+965109816n^{28}+4050441732n^{26}+8937136896n^{24}+11838786414n^{22}+16534395580n^{20}+11971009518n^{18}+9409389288n^{16}+3853491204n^{14}+973803384n^{12}-132081036n^{10}+119823968n^8+32622105n^6+1538106n^4-15551n^2+\color{red}{36})n}{36n^{38}-15551n^{36}+1538106n^{34}+32622105n^{32}+119823968n^{30}-132081036n^{28}+973803384n^{26}+3853491204n^{24}+9409389288n^{22}+11971009518n^{20}+16534395580n^{18}+11838786414n^{16}+8937136896n^{14}+4050441732n^{12}+965109816n^{10}+165024372n^8-1574764n^6+569433n^4+6234n^2+1)}$

I found an identity for $k=73$ using a known solution $(b,y)=(n, n(n^2-1)).$

$a=n,$

$b_7 = \frac{n(n^{48}+15704n^{46}+3430692n^{44}-57632376n^{42}+6702252562n^{40}+75079777160n^{38}+707080531380n^{36}+3414184912920n^{34}+7188385885663n^{32}+19104201262320n^{30}+33722429304776n^{28}+46168990682064n^{26}+57120264919228n^{24}+47015521855632n^{22}+37697060130056n^{20}+19106431585968n^{18}+8759831502031n^{16}+1465420875576n^{14}-8798478828n^{12}-70938314968n^{10}+2565235282n^8+603165288n^6+13866564n^4-54088n^2+\color{red}{49})}{(49n^{48}-54088n^{46}+13866564n^{44}+603165288n^{42}+2565235282n^{40}-70938314968n^{38}-8798478828n^{36}+1465420875576n^{34}+8759831502031n^{32}+19106431585968n^{30}+37697060130056n^{28}+47015521855632n^{26}+57120264919228n^{24}+46168990682064n^{22}+33722429304776n^{20}+19104201262320n^{18}+7188385885663n^{16}+3414184912920n^{14}+707080531380n^{12}+75079777160n^{10}+6702252562n^8-57632376n^6+3430692n^4+15704n^2+1)}$


II. New result for second family.

I found an identity for $k=43$ using a solution of degree $7$.

$a=n^3$,

$b = \frac{n(n^{30}+17n^{28}-18n^{26}+101n^{24}-172n^{22}+80n^{20}+282n^{18}-82n^{16}-244n^{14}+282n^{12}-28n^{10}-64n^8+101n^6+9n^4-10n^2+1)}{(n^{30}-10n^{28}+9n^{26}+101n^{24}-64n^{22}-28n^{20}+282n^{18}-244n^{16}-82n^{14}+282n^{12}+80n^{10}-172n^8+101n^6-18n^4+17n^2+1)}$

Tomita
  • 2,346
  • Great! Your solution also has a numerator and denominator whose coefficients are palindromes of each other. Plus, its square constant term $36$ fits in the pattern $4,9,16,25$ for my solutions. So there are now five solutions to $y^2=(n^3-b)(b^3-n).$ Using group law $mP$, is yours the point $m=6$? – Tito Piezas III Oct 05 '23 at 06:27
  • @Tito, The result was derived from group law $2P$ using $b= \frac{4 + n^2 + 10n^4 + n^6}{1 + 10n^2 + n^4 + 4n^6}.$ – Tomita Oct 05 '23 at 06:41
  • Tomita, I think I finally figured out the pattern. For odd $m=2n-1$, the $b_m$ generates identities of deg $k = 6(n^2-n)+1 = 1, 13, 37, 73, 121$ which explains why your $b_7$ yields deg-$73$ and my $b_9$ yields deg-$121$. For even $m=2n$, the $b_m$ generates identities of deg $k=6n^2+1 = 7,25,55,97$ which explains why my $b_4$ yields a deg-$25$ and your $b_6$ yields deg-$55$. If I'm right, the missing $b_8$ should yield your deg-$97$ identity. – Tito Piezas III Oct 06 '23 at 07:18
  • It's a very interesting sequence! But I can't figure out why. – Tomita Oct 06 '23 at 07:40
  • I need to check the relationship between my equation and $f(a,b)$. – Tomita Oct 06 '23 at 08:11
  • I found $b_8$! Thanks for providing the deg-$97$ $A$ term, as I derived the missing $b_8$ from it. I'll post it soon and credit it to us. I guess I was right about the pattern. – Tito Piezas III Oct 06 '23 at 13:42
  • I added $b_8$ to the MO post last night. Do you know if we can use the group law $2P, 3P,, \text{etc}$ to generate the $b_m$ for any desired $m$? – Tito Piezas III Oct 07 '23 at 03:05
  • I am not able to do so, but it seems that the numerator of $b_{2m-1}$ having a linear term that is $(2m-1)^2$ using group law $mP$ with $(b,y)=(n, n(n^2-1)).$ – Tomita Oct 07 '23 at 04:11
  • Kindly check out the answer in the MO post. A guy found $b_{10}$ and $b_{11}$ using group law, and claims one can do so for any $m$. How did he do it? (I asked him, but he hasn't replied yet.) – Tito Piezas III Oct 07 '23 at 07:24
  • He has already answered that. – Tomita Oct 07 '23 at 07:38
  • I deleted the deg-97 to prevent clutter in your answer, and since we already got $b_8$ from it anyway. – Tito Piezas III Oct 07 '23 at 10:57
2

I am trying to go structurally, details will be in there so that the yoga is transparent at lowest level. (Instead, a computer one-liner does often the job, and this is not only frustrating for the reader, but also for me when years later i do not recall the human argument and regret that typing in hurry.) Let us start.


Consider the curve defined by $$ v^2 = (u^3-A)(A^3 -u)\ , $$ first as a(n affine) curve in the variables $(u,v)$ over the field $F=\Bbb Q(A)$. So $A$ is felt like a fixed parameter. We bring it in the short Weierstrass form using the substitution $\displaystyle u=A^3-\frac 1{x_1}$, and $\displaystyle v=\frac 1{x_1^2}y_1$, this leads to $$ \begin{aligned} y_1^2 &= (A^3x_1 -1)^3 - Ax_1^3\ ,\\ y_1^2 &= \underbrace{(A^9-A)}_{:=c}x_1^3 - 3A^6x_1^2 +3A^3x_1 - 1\ ,\\ (cy_1)^2 &= (cx_1)^3 -3A^6(cx_1)^2 +3A^3c(cx_1)-c^2\ . \end{aligned} $$ The coefficient in $x_1^3$ on the R.H.S was $c=(A^9-A)$ in the second line. So we multiplied with $c^2$ on both sides, then used $y_2=cy_1$, $x_2=cx_1$. Here is the transformed equation: $$ y_2^2 = x_2^3 - 3A^6x_2^2 + 3A^3cx_2-c^2\ ,\qquad c=A^9-A\ . $$ The first two terms on the R.H.S. suggest the substitution $x=x_2-A^6$. We take $y=y_2$ and obtain a pretty simple form: $$ \bbox[yellow]{\qquad (E=E(A))\ :\qquad\qquad y^2 = x^3 -3A^4x-A^2(A^8+1)\ . \qquad } $$ When we pass from $(u,v)$ to $(x,y)$ the connecting formulas make $u$ depend only on $x$, and conversely. Explicitly: $$ \begin{aligned} x &= x_2-A^6=cx_1-A^6 =\frac c{A^3-u} - A^6\ ,\\ u &= A^3-\frac 1{x_1}=A^3-\frac c{x_2}=A^3-\frac c{x+A^6}\ . \end{aligned} $$ Do we have a quick guess of a point on $E(A)$? We indeed have such a guess if we specialize (base change, composition with $\Bbb Q[A]\to \Bbb Q[n]$, $A\to n^3$) to the case $A=n^3$. Then the factor $(u^3-A)=(u^3-n^3)$ has an easy point, $u=n$, and we obtain a point $H$ with $H_x=c/(A^3-n)-A^6=(A^9-A)/(A^3-n)-A^6=(n^{27}-n^3)/(n^9-n)-n^{18} =n^2(n^8+1)$. The corresponding $y$-value is $H_y=0$ (since the vanishing of $(u^3-A)$ makes $v$ vanish, which propagates to a vanishing $y$-value). Of course, the obtained point $$ H = H(n^3) = (H_x, H_y) = (n^2(n^8+1),\ 0) $$ on $E(n^3)$ is a two-torsion point. We need "more points", other (kind of) points.

What about the general case? It turns out that there is also a simple point on the $E(A)$ curve, namely $$ \bbox[yellow]{ \qquad G =G(A) = (G_x,G_y)=(A^4+A^2+1,\ A^6+A^4+A^2+1)\ . \qquad} $$ In general, when we specialize $A$ to a rational number, there is "only this" point (with its multiples) in the list of $\Bbb Q$-rational points. So we cannot expect a bigger rank for $E(A)$ over $\Bbb Q(A)$.

With the above arguments, it is easy to see why the cases

  • $A=n$ (where we have the point $G(A)$), and
  • $A=n^3$ (where we have linear combinations of the torsion point $H(n)$ and of $G(A)=G(n^3)$)

lead to solutions for the initial problem. In the context of the OP the following questions arise:

Question A: Which are explicitly the solutions? For $A=n$ and for $A=n^3$.

Question B: What about "specializations" of $A$ to rational numbers? In which cases do we get a "high rank"? Can we guess a formula for a generic family $A=\mathcal A(n)$ so that new parametric points appear?

Question C: What about "other generic specializations" $A=\mathcal A(n)$ possibly coming from B?


Answer to Question A, when parameter $A$ is $A=n$.

We start with $G$. Let $k$ be $1,2,3,4,\dots$ and compute $kG=(x_k,y_k)$. We need only $x_k$. Compute the corresponding $b_k =u_k:=A^3 - c/(x_k+A^6)$. It turns out that $b_k=b_k(n)$ is a function of the shape $b_k(n)=n\cdot\frac{g_k(n,1)}{g_k(1,n)}$.

Such a $b_k$ determines $v_k$ from $$ v_k^2 =(a^3-b_k)(b_k^3-a) \ . $$ We use it to compute $p,q,r,s$ as in the OP. We are interested in $p-q$, which is a polynomial in $n$. We compute it in $n=(N+1)/(N-1)$, and after clearing the denominator, a power of $(N-1)$, we obtain a polynomial $f_k=f_k(N)$. The postponed code gives the following tables for the $b$-coefficients, and polynomials $f$:

$k$ $g_k(n)$ which determines the solution $\displaystyle b_k(n) = n\cdot\frac{g_k(n,1)}{g_k(1,n)}$
$1$ $1$
$2$ $n^{6} + 10n^{4} + n^{2} + \color{red}{4}$
$3$ $n^{8} + 100n^{6} + 190n^{4} - 44n^{2} + \color{red}{9}$
$4$ $n^{18} + 552n^{16} + 5820n^{14} + 6568n^{12} + 22758n^{10} + 10632n^{8} + 15100n^{6} + 4632n^{4} - 543n^{2} + \color{red}{16}$
$5$ $n^{24} + 2092n^{22} + 69418n^{20} + 45836n^{18} + 2404327n^{16} + 3492760n^{14} + 5807660n^{12} + 2379064n^{10} + 1486687n^{8} + 979388n^{6} + 113482n^{4} - 3524n^{2} + \color{red}{25}$
$6$ $n^{38} + 6234n^{36} + 569433n^{34} - 1574764n^{32} + 165024372n^{30} + 965109816n^{28} + 4050441732n^{26} + 8937136896n^{24} + 11838786414n^{22} + 16534395580n^{20} + 11971009518n^{18} + 9409389288n^{16} + 3853491204n^{14} + 973803384n^{12} - 132081036n^{10} + 119823968n^{8} + 32622105n^{6} + 1538106n^{4} - 15551n^{2} + \color{red}{36}$
$7$ $\small n^{48} + 15704n^{46} + 3430692n^{44} - 57632376n^{42} + 6702252562n^{40} + 75079777160n^{38} + 707080531380n^{36} + 3414184912920n^{34} + 7188385885663n^{32} + 19104201262320n^{30} + 33722429304776n^{28} + 46168990682064n^{26} + 57120264919228n^{24} + 47015521855632n^{22} + 37697060130056n^{20} + 19106431585968n^{18} + 8759831502031n^{16} + 1465420875576n^{14} - 8798478828n^{12} - 70938314968n^{10} + 2565235282n^{8} + 603165288n^{6} + 13866564n^{4} - 54088n^{2} + \color{red}{49}$
$8$ $\small n^{66} + 34976n^{64} + 16598640n^{62} - 802597088n^{60} + 169138412280n^{58} + 3341483957216n^{56} + 68921349810256n^{54} + 698110524939936n^{52} + 2371147488742812n^{50} + 11300859600110560n^{48} + 58394313096107888n^{46} + 197366608205845152n^{44} + 540780268830710728n^{42} + 1119960092282981472n^{40} + 1898187664589514704n^{38} + 2559531081677676064n^{36} + 3024275414914466502n^{34} + 2862767624014048416n^{32} + 2438532655839868624n^{30} + 1699152015688998752n^{28} + 1059537396760387272n^{26} + 565197867108725152n^{24} + 262189050432532080n^{22} + 108715494169641440n^{20} + 30950084935006876n^{18} + 6235144197369504n^{16} + 533059363441488n^{14} + 1644147453664n^{12} - 5988361467400n^{10} + 25016401440n^{8} + 7908319600n^{6} + 91670880n^{4} - 158335n^{2} + \color{red}{64}$
$9$ $\tiny n^{80} + 70888n^{78} + 67097980n^{76} - 7601467144n^{74} + 2931242386030n^{72} + 88440504999640n^{70} + 3884690979741692n^{68} + 68841496974471880n^{66} + 159626691817966525n^{64} + 2343223081001783712n^{62} + 42614192414467088816n^{60} + 299652084748204903904n^{58} + 1516563414177942983208n^{56} + 5723686723196286338272n^{54} + 17178808479864835237360n^{52} + 38818758949934604483488n^{50} + 74255179165990324638802n^{48} + 117574894628368241794864n^{46} + 160579318360111593624264n^{44} + 186242485219018870892816n^{42} + 188963554058602494368852n^{40} + 161721875649245818566864n^{38} + 120310095858827397125128n^{36} + 74373521995862636237296n^{34} + 37939061001155279992178n^{32} + 16007166226577445125920n^{30} + 4989911070816764106352n^{28} + 1531812067426248122976n^{26} + 521545149005486710568n^{24} + 234981869722959440480n^{22} + 81169043718961628976n^{20} + 14686441538438168608n^{18} + 1804737030046873645n^{16} + 96071436547023656n^{14} + 1757845056647068n^{12} - 269073613222472n^{10} - 336916143634n^{8} + 77282464024n^{6} + 482840284n^{4} - 407672n^{2} + \color{red}{81}$
$10$ $\tiny n^{102} + 133370n^{100} + 235431945n^{98} - 53960558412n^{96} + 37353015835420n^{94} + 1715285459850920n^{92} + 154018873425409932n^{90} + 4528460790052228992n^{88} - 10283455511101671494n^{86} + 424389497721502142548n^{84} + 21198683938074571284634n^{82} + 261033154046082383403576n^{80} + 2270064982003402818480028n^{78} + 15317757618354379225002184n^{76} + 84218098664615395090065004n^{74} + 345159573021345914583822624n^{72} + 1129951624304798874757797847n^{70} + 3131305476125014022272232326n^{68} + 7635830330877839055871498111n^{66} + 16446578072159313295259200380n^{64} + 31726949494863946397847804664n^{62} + 54769575000613973884608095824n^{60} + 84669779180983943334398860888n^{58} + 117776609406232459328970393984n^{56} + 146631440164415842271230869964n^{54} + 164805637943248865339648114968n^{52} + 165741869909257891183520432332n^{50} + 150306415334481456543373768912n^{48} + 121703129810992794049207385688n^{46} + 88124674373459097943143748048n^{44} + 56481814917222827023209545464n^{42} + 31767118902941960531617938880n^{40} + 15407922021660173371718373183n^{38} + 6306682290739982829447936934n^{36} + 2063724483753154237664688919n^{34} + 509509860910406920886925292n^{32} + 72423661674195782496620652n^{30} - 2085965471015373600109304n^{28} - 3102450405774431050519652n^{26} - 511897235486864832178688n^{24} + 72673456634449307534874n^{22} + 33312001105932688908244n^{20} + 3846262561102571988538n^{18} + 294258428803841831032n^{16} + 9380206793092459404n^{14} + 156074955019903848n^{12} - 7836007732260580n^{10} - 12400565501408n^{8} + 607383986505n^{6} + 2125016730n^{4} - 948799n^{2} + \color{red}{100}$

and respectively

$k$ Polynomial $f_k$ obtained from $a=a(n)=n$, $b=b(n)=b_k(n)$ above, and $v$ with $v^2=(a^3 - b)(b^3-a)$
$1$ $1$
$2$ $n^{7} + n^{5} - 2n^{3} - 3n^{2} + n$
$3$ $n^{13} + n^{12} + 2n^{11} - 4n^{10} - 3n^{9} + 3n^{8} + 7n^{7} + 4n^{6} - 12n^{5} - 6n^{4} + 5n^{3} - n^{2} + n + 1$
$4$ $n^{24} + 3n^{23} + 10n^{22} + 9n^{21} + 4n^{20} - 9n^{19} - 97n^{18} + 57n^{17} + 136n^{16} - 207n^{15} - 128n^{14} + 333n^{13} + 179n^{12} - 213n^{11} - 134n^{10} - 36n^{9} + 43n^{8} + 63n^{7} - 25n^{6} + 3n^{5} - 2n^{4} + 7n^{2} + 1$
$5$ $n^{37} + 3n^{35} - 12n^{33} - 6n^{32} - 25n^{31} - 129n^{30} - 3n^{29} + 36n^{28} + 597n^{27} + 996n^{26} - 2091n^{25} - 1443n^{24} + 2391n^{23} - 876n^{22} + 810n^{21} + 5040n^{20} - 3683n^{19} - 7572n^{18} + 1833n^{17} + 6063n^{16} + 774n^{15} - 2493n^{14} - 417n^{13} + 231n^{12} - 195n^{11} + 204n^{10} - 96n^{9} - 57n^{8} + 140n^{7} + 36n^{6} - 27n^{5} - 21n^{4} - 3n^{3} - 6n^{2} + n$
$6$ $\small n^{55} + n^{54} + 12n^{53} + 33n^{51} - 57n^{50} - 42n^{49} + 78n^{48} - 54n^{47} + 900n^{46} + 561n^{45} - 1989n^{44} - 3696n^{43} - 2229n^{42} - 5535n^{41} - 5217n^{40} + 68337n^{39} + 37017n^{38} - 150449n^{37} - 26405n^{36} + 116829n^{35} - 18705n^{34} + 15855n^{33} - 54738n^{32} - 113172n^{31} + 144816n^{30} + 242436n^{29} + 97650n^{28} - 411291n^{27} - 556077n^{26} + 417051n^{25} + 675039n^{24} - 298188n^{23} - 368781n^{22} + 217650n^{21} + 82116n^{20} - 143540n^{19} - 19496n^{18} + 60858n^{17} + 29538n^{16} - 15852n^{15} - 15534n^{14} - 270n^{13} + 1197n^{12} + 3699n^{11} + 1149n^{10} - 1035n^{9} - 81n^{8} - 288n^{7} - 168n^{6} + 66n^{5} - 24n^{4} + 18n^{3} + 6n^{2} + n + 1$
$7$ $\tiny n^{72} + 6n^{71} + 21n^{70} + 57n^{69} + 99n^{68} + 108n^{67} - 404n^{66} + 24n^{65} - 2472n^{64} + 360n^{63} + 6612n^{62} - 9987n^{61} + 13420n^{60} + 3543n^{59} - 74715n^{58} + 198468n^{57} + 174426n^{56} - 589023n^{55} - 122065n^{54} - 286626n^{53} - 1051764n^{52} + 4220589n^{51} + 4661865n^{50} - 7808517n^{49} - 11088089n^{48} + 3529014n^{47} + 19401177n^{46} + 8186328n^{45} - 25427109n^{44} - 17906667n^{43} + 23726308n^{42} + 23334441n^{41} - 15808299n^{40} - 35153883n^{39} + 3360372n^{38} + 60606342n^{37} + 20504051n^{36} - 86424528n^{35} - 47910273n^{34} + 89016168n^{33} + 49873428n^{32} - 62965773n^{31} - 21332798n^{30} + 26764431n^{29} - 6849006n^{28} - 1507041n^{27} + 14365122n^{26} - 6189936n^{25} - 9469508n^{24} + 3580719n^{23} + 4021995n^{22} - 237888n^{21} - 1003026n^{20} - 550458n^{19} - 83500n^{18} + 162465n^{17} + 138423n^{16} + 57486n^{15} - 20772n^{14} - 27585n^{13} - 4178n^{12} - 5079n^{11} + 1173n^{10} + 2052n^{9} - 444n^{8} + 381n^{7} - 131n^{6} + 6n^{5} + 48n^{4} + 15n^{2} + 1$

Used code (with True, False as shown to get the $f$-polynomials, respectively with False, True instead to get the $g$-polynomials):

def E(A, field=QQ):
    return EllipticCurve(field, [-3*A^4, -A^10 - A^2])

R0.<n> = PolynomialRing(ZZ) F = FractionField(R0) E1 = E(n, field=F) G = E1.point([n^4 + n^2 + 1, n^6 + n^4 + n^2 + 1])

for k in [1..7]: a = n x = (kG)[0] u = a^3 - (a^9 - a)/(x + a^6) b = u v = sqrt( (a^3 - b)(b^3 - a) ) # print(f"a = {a}\nb = {b}\nv = {v}") p, q, r, s = a^3 - b, av, b(a^3 - b), v den = lcm([ expr.denominator() for expr in (p, q, r, s)]) p, q, r, s = pden, qden, rden, sden g = gcd([p, q, r, s]) p, q, r, s = R0(p/g), R0(q/g), R0(r/g), R0(s/g)

# print(f&quot;Check (p+q)^4 + (r-s)^4 == (r+s)^4 + (p-q)^4: &quot;
#       f&quot;{(p+q)^4 + (r-s)^4 == (r+s)^4 + (p-q)^4}&quot;)

f = (p - q).subs(n=(n + 1)/(n - 1)).numerator()
f = R0(f / f.content() )

if True:    # we show f :: set it to False if not wanted
    # print(f&quot;f = {f}\n\n&quot;)
    print(f&quot;| <span class="math-container">${k}$</span> | <span class="math-container">${latex(f)}$</span> |&quot;)

if False:   # we show g :: set it to False if not wanted
    g = (b/n).numerator()
    if b == n * g(n) / g(1/n) / n^g.degree():
        # print(f&quot;g = {g}\n\n&quot;)
        print(f&quot;| <span class="math-container">${k}$</span> | <span class="math-container">${latex(g)}$</span> |&quot;)


... to be continued ...

(There is a maximal amout of characters that can be used in an answer, and the tables take them all...)

dan_fulea
  • 32,856
  • I have the answer to Question B. Instead of $A=n$, or $A=n^3$, there is also a 3rd family of $A$ that yields an identity not in the 1st or 2nd families. This is given by, $$A=\frac{n(1-2n^2+n^4+n^6)}{1+n^2-2n^4+n^6}$$ A first solution to $(A^3-b)(b^3-A) = y^2$ is $b_1 = n^3$. A third solution is, $$\small{b_3=\frac{n(1+n^2-20n^4-7n^6 + 199n^8 - 314n^{10} + 80n^{12} + 163n^{14} - 116n^{16} - 7n^{18} + 25n^{20} + 10n^{22} + n^{24})}{1 + 10n^2 + 25n^4 - 7n^6 - 116n^8 + 163n^{10} + 80n^{12} - 314n^{14} + 199n^{16} - 7n^{18} - 20n^{20} + n^{22} + n^{24}}}$$ This will yield a deg-$31$ identity. – Tito Piezas III Oct 07 '23 at 16:33
  • Yes, this is an important observation, i tried to use it in a structural manner in my answer in as follows. We start with the equation $v^2=(a^3-b)(b^3-a)$. It can be seen as a family of elliptic curves $E(A)$, where the parameter $A$ takes the value $a$, and we have rational points over $\Bbb Q(A)$, one of them is $G(A)$, but we may take some other point $P=P(n)$. It can be seen as "an other" family $E'(B)$. And we would like to have specializations of the 1 parameter that still lead to "known" points for the other. Yes, take $P=P(n)$ as a special family! But i can't use better+effectively. – dan_fulea Oct 07 '23 at 18:09
2

... continued answer (please search around for the beginning, i had to split since there is a maximal bound of the number of characters in each answer)...


Answer to Question A, when parameter $A$ is $A=n^3$.

Similar code produces the tables for $f$-polynomials and $g$-polynomials, with an additional check. Here we use the points $kG(n^3) + jH(n^3)$. We obtain (arguably) "nothing new" when we take $j=0$.

In the sense that starting with a "primitive" solution $F(\alpha,\beta)$ we can do a lot of things to obtain without work other solutions. We can plug in instead of $\alpha,\beta$ any polynomial expressions in $\alpha,\beta$. We can use a Möbius transform. Also, our $f$ was obtained from a solution $a=a(n)$, $b=b(n)$, $v=v(n)$. Some factors were simplified. We can simplify less, or use our own factors. It is hard to say if some solution in some degree does not come from a primitive one, by using operations of the mentioned kind.

To keep the tables short, i will use $kG+jH$ for $k\le 6$ and $j=0,1$. But of course, the code produces a correct, checked output also beyond.

$k$ $j$ $g_k(n)$ which determines the solution $\displaystyle b_k(n) = n\cdot\frac{G_k(n,1)}{G_k(1,n)}$ with $G_k=g_k^{\text{homogenized}}$
$1$ $0$ $n^{2}$
$1$ $1$ $n^{6} + n^{4} - 2n^{2} + 1$
$2$ $0$ $n^{20} + 10n^{14} + n^{8} + 4n^{2}$
$2$ $1$ $n^{12} + 8n^{10} + 10n^{6} - 4n^{2} + 1$
$3$ $0$ $n^{26} + 100n^{20} + 190n^{14} - 44n^{8} + 9n^{2}$
$3$ $1$ $n^{30} + 17n^{28} - 18n^{26} + 101n^{24} - 172n^{22} + 80n^{20} + 282n^{18} - 82n^{16} - 244n^{14} + 282n^{12} - 28n^{10} - 64n^{8} + 101n^{6} + 9n^{4} - 10n^{2} + 1$
$4$ $0$ $\small n^{56} + 552n^{50} + 5820n^{44} + 6568n^{38} + 22758n^{32} + 10632n^{26} + 15100n^{20} + 4632n^{14} - 543n^{8} + 16n^{2}$
$4$ $1$ $n^{48} + 32n^{46} + 552n^{42} - 1088n^{40} - 16n^{38} + 5820n^{36} + 8160n^{34} + 544n^{32} + 6552n^{30} + 18560n^{28} - 4080n^{26} + 23302n^{24} + 8160n^{22} - 9280n^{20} + 6552n^{18} - 1088n^{16} - 4080n^{14} + 5820n^{12} + 32n^{10} + 544n^{8} + 552n^{6} - 16n^{2} + 1$
$5$ $0$ $\small n^{74} + 2092n^{68} + 69418n^{62} + 45836n^{56} + 2404327n^{50} + 3492760n^{44} + 5807660n^{38} + 2379064n^{32} + 1486687n^{26} + 979388n^{20} + 113482n^{14} - 3524n^{8} + 25n^{2}$
$5$ $1$ $\small n^{78} + 49n^{76} - 50n^{74} + 2093n^{72} - 9092n^{70} + 7024n^{68} + 71486n^{66} + 146362n^{64} - 221372n^{62} + 120846n^{60} + 1989884n^{58} - 1997248n^{56} + 2411691n^{54} + 2513095n^{52} - 3945398n^{50} + 4925063n^{48} + 1374136n^{46} - 4812512n^{44} + 9246036n^{42} + 3689036n^{40} - 10556008n^{38} + 9246036n^{36} + 2487832n^{34} - 5926208n^{32} + 4925063n^{30} + 3430735n^{28} - 4863038n^{26} + 2411691n^{24} + 1056332n^{22} - 1063696n^{20} + 120846n^{18} + 102298n^{16} - 177308n^{14} + 71486n^{12} - 3476n^{10} + 1408n^{8} + 2093n^{6} + 25n^{4} - 26n^{2} + 1$
$6$ $0$ $\tiny n^{116} + 6234n^{110} + 569433n^{104} - 1574764n^{98} + 165024372n^{92} + 965109816n^{86} + 4050441732n^{80} + 8937136896n^{74} + 11838786414n^{68} + 16534395580n^{62} + 11971009518n^{56} + 9409389288n^{50} + 3853491204n^{44} + 973803384n^{38} - 132081036n^{32} + 119823968n^{26} + 32622105n^{20} + 1538106n^{14} - 15551n^{8} + 36n^{2}$
$6$ $1$ $\tiny n^{108} + 72n^{106} + 6234n^{102} - 31104n^{100} - 36n^{98} + 569433n^{96} + 3063744n^{94} + 15552n^{92} - 1574800n^{90} + 64105344n^{88} - 1531872n^{86} + 165039924n^{84} + 242797536n^{82} - 32052672n^{80} + 963577944n^{78} - 594241920n^{76} - 121398768n^{74} + 4018389060n^{72} + 20450880n^{70} + 297120960n^{68} + 8815738128n^{66} - 329795712n^{64} - 10225440n^{62} + 12135907374n^{60} + 1187302320n^{58} + 164897856n^{56} + 16524170140n^{54} - 329795712n^{52} - 593651160n^{50} + 12135907374n^{48} + 20450880n^{46} + 164897856n^{44} + 8815738128n^{42} - 594241920n^{40} - 10225440n^{38} + 4018389060n^{36} + 242797536n^{34} + 297120960n^{32} + 963577944n^{30} + 64105344n^{28} - 121398768n^{26} + 165039924n^{24} + 3063744n^{22} - 32052672n^{20} - 1574800n^{18} - 31104n^{16} - 1531872n^{14} + 569433n^{12} + 72n^{10} + 15552n^{8} + 6234n^{6} - 36n^{2} + 1$
$7$ $0$ $\tiny n^{146} + 15704n^{140} + 3430692n^{134} - 57632376n^{128} + 6702252562n^{122} + 75079777160n^{116} + 707080531380n^{110} + 3414184912920n^{104} + 7188385885663n^{98} + 19104201262320n^{92} + 33722429304776n^{86} + 46168990682064n^{80} + 57120264919228n^{74} + 47015521855632n^{68} + 37697060130056n^{62} + 19106431585968n^{56} + 8759831502031n^{50} + 1465420875576n^{44} - 8798478828n^{38} - 70938314968n^{32} + 2565235282n^{26} + 603165288n^{20} + 13866564n^{14} - 54088n^{8} + 49n^{2}$
$7$ $1$ $\tiny n^{150} + 97n^{148} - 98n^{146} + 15705n^{144} - 123784n^{142} + 108128n^{140} + 3446348n^{138} + 24162948n^{136} - 27663384n^{134} - 54131940n^{132} + 1284695208n^{130} - 1216696704n^{128} + 6634254058n^{126} - 229454414n^{124} - 5801634356n^{122} + 81110865930n^{120} - 223888114072n^{118} + 145342483424n^{116} + 785626162028n^{114} - 1023645380268n^{112} + 167080903272n^{110} + 4270749389916n^{108} - 2214069073416n^{106} - 2065478795328n^{104} + 11467933754407n^{102} + 4703023132063n^{100} - 14705536010894n^{98} + 29106714141151n^{96} + 16623299156016n^{94} - 36970181795136n^{92} + 54069311943896n^{90} + 39190788849032n^{88} - 74153669206960n^{86} + 81131871039992n^{84} + 53330412573456n^{82} - 96765223483392n^{80} + 100555075829164n^{78} + 64281686810620n^{76} - 117821240784152n^{74} + 100555075829164n^{72} + 52483881399888n^{70} - 95918692309824n^{68} + 81131871039992n^{66} + 35216158023752n^{64} - 70179038381680n^{62} + 54069311943896n^{60} + 16621068832368n^{58} - 36967951471488n^{56} + 29106714141151n^{54} + 3131577515695n^{52} - 13134090394526n^{50} + 11467933754407n^{48} - 265305036072n^{46} - 4014242832672n^{44} + 4270749389916n^{42} - 307766370060n^{40} - 548798106936n^{38} + 785626162028n^{36} - 77870021944n^{34} - 675608704n^{32} + 81110865930n^{30} + 3907562866n^{28} - 9938651636n^{26} + 6634254058n^{24} + 623897544n^{22} - 555899040n^{20} - 54131940n^{18} + 13727076n^{16} - 17227512n^{14} + 3446348n^{12} - 53992n^{10} + 38336n^{8} + 15705n^{6} + 49n^{4} - 50n^{2} + 1$

(for instance the line for $(k,j)=(2,0)$ produces no new result, it is the same as the one for $k=0$ in the $A=n$ case...)

and

$k$ $j$ Polynomial $f_k$ obtained from $a=n^3$, $b=b_k(n)$ as above, and $v$ with $v^2=(a^3 - b)(b^3-a)$
$1$ $0$ $n^{3} - 1$
$1$ $1$ $n^{7} + n^{5} - 2n^{3} - 3n^{2} + n$
$2$ $0$ $2n^{21} - n^{18} + 20n^{15} + 17n^{12} + 2n^{9} + 17n^{6} + 8n^{3} - 1$
$2$ $1$ $n^{19} + 6n^{17} - 18n^{15} - 6n^{14} - 5n^{13} - 12n^{12} - 12n^{11} - 36n^{10} - 24n^{9} + 12n^{8} + 19n^{7} - 36n^{6} + 6n^{5} - 12n^{4} - 6n^{3} - 6n^{2} + n$
$3$ $0$ $3n^{39} - n^{36} + 294n^{33} + 214n^{30} - 27n^{27} + 2481n^{24} - 972n^{21} + 2804n^{18} + 861n^{15} + 2481n^{12} - 186n^{9} + 214n^{6} + 27n^{3} - 1$
$3$ $1$ $n^{43} + 13n^{41} - 98n^{39} - 9n^{38} - 33n^{37} - 36n^{36} - 406n^{35} - 252n^{34} - 616n^{33} + 486n^{32} + 1973n^{31} - 1260n^{30} - 2949n^{29} - 180n^{28} - 1818n^{27} + 441n^{26} + 1507n^{25} + 1944n^{24} - 1940n^{23} - 216n^{22} - 2912n^{21} - 1836n^{20} + 3667n^{19} - 216n^{18} - 2061n^{17} + 1944n^{16} - 2166n^{15} + 441n^{14} + 893n^{13} - 180n^{12} - 886n^{11} - 1260n^{10} - 568n^{9} + 486n^{8} + 183n^{7} - 252n^{6} + 37n^{5} - 36n^{4} - 14n^{3} - 9n^{2} + n$
$4$ $0$ $\small 4n^{75} - n^{72} + 2140n^{69} + 1220n^{66} - 14324n^{63} + 69902n^{60} - 407020n^{57} + 389428n^{54} - 749144n^{51} + 1480145n^{48} - 1928360n^{45} + 960008n^{42} - 2183848n^{39} + 2587204n^{36} - 1640216n^{33} + 960008n^{30} - 1301420n^{27} + 1480145n^{24} - 217588n^{21} + 389428n^{18} + 54364n^{15} + 69902n^{12} - 3260n^{9} + 1220n^{6} + 64n^{3} - 1$
$4$ $1$ $\small n^{73} + 24n^{71} - 288n^{69} - 12n^{68} - 452n^{67} - 96n^{66} - 4944n^{65} - 1152n^{64} - 11544n^{63} + 2796n^{62} + 51634n^{61} - 13536n^{60} - 101064n^{59} - 3552n^{58} - 169008n^{57} - 19236n^{56} - 87604n^{55} + 225792n^{54} - 450912n^{53} + 298656n^{52} - 1088568n^{51} - 486012n^{50} + 82735n^{49} + 681600n^{48} + 412080n^{47} + 2277504n^{46} - 2893056n^{45} - 27192n^{44} + 496120n^{43} + 3033792n^{42} + 51360n^{41} + 2734464n^{40} - 2550192n^{39} + 529656n^{38} + 1487420n^{37} + 3349440n^{36} + 619440n^{35} + 3349440n^{34} - 3687840n^{33} + 529656n^{32} + 886264n^{31} + 2734464n^{30} - 388992n^{29} + 3033792n^{28} - 1559472n^{27} - 27192n^{26} + 973615n^{25} + 2277504n^{24} + 85560n^{23} + 681600n^{22} - 532320n^{21} - 486012n^{20} + 304076n^{19} + 298656n^{18} - 201936n^{17} + 225792n^{16} - 86328n^{15} - 19236n^{14} - 590n^{13} - 3552n^{12} - 20712n^{11} - 13536n^{10} - 4272n^{9} + 2796n^{8} + 1084n^{7} - 1152n^{6} + 96n^{5} - 96n^{4} - 24n^{3} - 12n^{2} + n$
$5$ $0$ $\tiny 5n^{111} - n^{108} + 10110n^{105} + 4674n^{102} - 386571n^{99} + 955239n^{96} - 27124016n^{93} + 13387696n^{90} - 106454652n^{87} + 179189772n^{84} - 927672120n^{81} - 371803464n^{78} - 4830485964n^{75} - 1439960196n^{72} - 8378654544n^{69} - 8973677616n^{66} - 14529143322n^{63} - 14630108910n^{60} - 12746371084n^{57} - 18275451124n^{54} - 13361170698n^{51} - 14630108910n^{48} - 7681208400n^{45} - 8973677616n^{42} - 3614271276n^{39} - 1439960196n^{36} - 1989116472n^{33} - 371803464n^{30} - 499480476n^{27} + 179189772n^{24} - 29685296n^{21} + 13387696n^{18} + 1764285n^{15} + 955239n^{12} - 26370n^{9} + 4674n^{6} + 125n^{3} - 1$
$5$ $1$ $\tiny n^{115} + 37n^{113} - 722n^{111} - 15n^{110} - 1397n^{109} - 180n^{108} - 29154n^{107} - 3420n^{106} - 81624n^{105} + 15390n^{104} + 824571n^{103} - 104220n^{102} - 2356107n^{101} + 54540n^{100} - 4133526n^{99} - 504567n^{98} - 7654711n^{97} + 7714656n^{96} - 16672048n^{95} + 10031904n^{94} - 105477376n^{93} - 40215216n^{92} + 38806628n^{91} + 62582688n^{90} + 356572164n^{89} + 332798688n^{88} - 1286314440n^{87} - 70838604n^{86} + 728490828n^{85} + 698824080n^{84} + 2101070664n^{83} + 805181616n^{82} - 3122294112n^{81} + 1824499656n^{80} + 3366922620n^{79} + 2196105264n^{78} + 6549997620n^{77} - 208278000n^{76} - 8751182136n^{75} + 10896531012n^{74} + 10661862996n^{73} - 191554272n^{72} + 7588831920n^{71} + 3353042016n^{70} - 16148171136n^{69} + 26785036080n^{68} + 24335459454n^{67} - 875999520n^{66} + 11006992806n^{65} - 3222656352n^{64} - 23923543836n^{63} + 39424474638n^{62} + 28089253738n^{61} - 2576683512n^{60} + 13782166708n^{59} - 391055976n^{58} - 28464292304n^{57} + 48520433460n^{56} + 33427945594n^{55} - 391055976n^{54} + 9202822134n^{53} - 2576683512n^{52} - 25471019892n^{51} + 39424474638n^{50} + 20754284862n^{49} - 3222656352n^{48} + 11000754096n^{47} - 875999520n^{46} - 17801490432n^{45} + 26785036080n^{44} + 12288897684n^{43} + 3353042016n^{42} + 4349530452n^{41} - 191554272n^{40} - 8916475560n^{39} + 10896531012n^{38} + 3198889308n^{37} - 208278000n^{36} + 2410431048n^{35} + 2196105264n^{34} - 3035063520n^{33} + 1824499656n^{32} + 400913004n^{31} + 805181616n^{30} + 320858724n^{29} + 698824080n^{28} - 365189784n^{27} - 70838604n^{26} + 134529188n^{25} + 332798688n^{24} + 73259984n^{23} + 62582688n^{22} - 42579328n^{21} - 40215216n^{20} + 19483529n^{19} + 10031904n^{18} - 4569891n^{17} + 7714656n^{16} - 1616802n^{15} - 504567n^{14} + 41571n^{13} + 54540n^{12} - 184674n^{11} - 104220n^{10} - 26904n^{9} + 15390n^{8} + 4003n^{7} - 3420n^{6} + 253n^{5} - 180n^{4} - 38n^{3} - 15n^{2} + n$
$6$ $0$ $\tiny 6n^{165} - n^{162} + 36108n^{159} + 13977n^{156} - 4674978n^{153} + 8298837n^{150} - 824175096n^{147} + 259552995n^{144} - 3907101468n^{141} + 11118044610n^{138} - 182026391688n^{135} - 167756888178n^{132} - 3274347963132n^{129} - 1520492261730n^{126} - 15152876115936n^{123} - 18470188571598n^{120} - 49509246929070n^{117} - 106779527182107n^{114} - 72349277452140n^{111} - 373599512128477n^{108} - 37946306899062n^{105} - 900567003891249n^{102} - 11153463766008n^{99} - 1792349450153127n^{96} + 132309743405400n^{93} - 2586188158430484n^{90} + 105690483035664n^{87} - 3227568551144460n^{84} + 142983549120600n^{81} - 3227568551144460n^{78} - 9630724747776n^{75} - 2586188158430484n^{72} - 31227836320566n^{69} - 1792349450153127n^{66} - 85600174473228n^{63} - 900567003891249n^{60} - 47726439661038n^{57} - 373599512128477n^{54} - 17641703806440n^{51} - 106779527182107n^{48} + 1929455404164n^{45} - 18470188571598n^{42} - 911968749576n^{39} - 1520492261730n^{36} - 539847092892n^{33} - 167756888178n^{30} - 60677712672n^{27} + 11118044610n^{24} - 1605768930n^{21} + 259552995n^{18} + 28939500n^{15} + 8298837n^{12} - 139962n^{9} + 13977n^{6} + 216n^{3} - 1$
$6$ $1$ $\tiny n^{163} + 54n^{161} - 1458n^{159} - 18n^{158} - 5229n^{157} - 324n^{156} - 130572n^{155} - 8748n^{154} - 573048n^{153} + 49140n^{152} + 6976143n^{151} - 498636n^{150} - 23444802n^{149} + 209628n^{148} - 84405726n^{147} - 6270426n^{146} - 363490899n^{145} + 122297040n^{144} - 578441520n^{143} + 303905520n^{142} - 5532784704n^{141} - 972576432n^{140} + 398044206n^{139} + 1012563504n^{138} + 45907613892n^{137} + 20375806608n^{136} - 137391796188n^{135} + 19196435412n^{134} + 78622211754n^{133} + 89670921048n^{132} + 566434662120n^{131} + 82414750536n^{130} - 80159355216n^{129} + 693489798504n^{128} + 1960823702922n^{127} + 915872402376n^{126} + 4638730009860n^{125} - 767286798120n^{124} + 1123704146700n^{123} + 8162433304308n^{122} + 17426186773470n^{121} + 11169411408n^{120} + 7571866083984n^{119} + 1984669712496n^{118} - 5263789218048n^{117} + 52063186875408n^{116} + 109710831178251n^{115} - 2022480178128n^{114} - 6150243333726n^{113} + 4813003298448n^{112} - 15562834222662n^{111} + 136148846855562n^{110} + 359262505440145n^{109} - 15776621125596n^{108} - 14947974789300n^{107} - 20569794758964n^{106} + 16128768933720n^{105} + 208358709948108n^{104} + 945116539368237n^{103} + 41556074545836n^{102} - 27378927840150n^{101} + 14004725123268n^{100} + 23017956151878n^{99} + 185240960999586n^{98} + 1713427014754503n^{97} - 42816539354976n^{96} - 1123478976096n^{95} + 10703037542112n^{94} + 2102750817408n^{93} - 78882349460832n^{92} + 2671610316398580n^{91} + 46709085266784n^{90} + 60338280404376n^{89} - 15381066513888n^{88} + 19836023097240n^{87} - 276074828634888n^{86} + 3172667153227164n^{85} - 51190758653616n^{84} - 29229905762640n^{83} + 27633010134384n^{82} - 49634072805600n^{81} - 471457334646864n^{80} + 3233518254314172n^{79} + 27633010134384n^{78} + 69253055706456n^{77} - 51190758653616n^{76} - 24632610749496n^{75} - 276074828634888n^{74} + 2619462866520756n^{73} - 15381066513888n^{72} - 66796283191392n^{71} + 46709085266784n^{70} + 12953593799424n^{69} - 78882349460832n^{68} + 1754494120181511n^{67} + 10703037542112n^{66} - 58326943590n^{65} - 42816539354976n^{64} - 1646501791518n^{63} + 185240960999586n^{62} + 914790447128709n^{61} + 14004725123268n^{60} - 10178166439476n^{59} + 41556074545836n^{58} + 32686768440504n^{57} + 208358709948108n^{56} + 378114860427769n^{55} - 20569794758964n^{54} + 7016849563698n^{53} - 15776621125596n^{52} + 2039698883214n^{51} + 136148846855562n^{50} + 101299889308779n^{49} + 4813003298448n^{48} + 6912868753296n^{47} - 2022480178128n^{46} - 8140269137472n^{45} + 52063186875408n^{44} + 19640997505278n^{43} + 1984669712496n^{42} - 731175099804n^{41} + 11169411408n^{40} - 4017376671228n^{39} + 8162433304308n^{38} + 1782726565146n^{37} - 767286798120n^{36} + 227707526376n^{35} + 915872402376n^{34} - 746951980752n^{33} + 693489798504n^{32} + 24862357530n^{31} + 82414750536n^{30} + 19161470052n^{29} + 89670921048n^{28} - 20562826644n^{27} + 19196435412n^{26} + 9001667214n^{25} + 20375806608n^{24} + 4306552272n^{23} + 1012563504n^{22} - 1106030592n^{21} - 972576432n^{20} + 449653197n^{19} + 303905520n^{18} - 82727730n^{17} + 122297040n^{16} - 19814058n^{15} - 6270426n^{14} - 739593n^{13} + 209628n^{12} - 1226124n^{11} - 498636n^{10} - 105624n^{9} + 49140n^{8} + 12267n^{7} - 8748n^{6} + 486n^{5} - 324n^{4} - 54n^{3} - 18n^{2} + n$
$7$ $0$ $\tiny 7n^{219} - n^{216} + 106068n^{213} + 35268n^{210} - 36681366n^{207} + 52083786n^{204} - 14882514188n^{201} + 3320546180n^{198} + 5230427391n^{195} + 395751972327n^{192} - 19448287215072n^{189} - 21640544005728n^{186} - 829109912989328n^{183} - 282571560087184n^{180} - 6105712140674400n^{177} - 9326892512301792n^{174} - 28339046510383188n^{171} - 124533892690899060n^{168} + 138203765744892272n^{165} - 929843481131980112n^{162} + 2276825051545128984n^{159} - 4471733528882428008n^{156} + 10598935646776627824n^{153} - 18831830242753062096n^{150} + 35492404154683555804n^{147} - 57123279707358859780n^{144} + 90714271101777486432n^{141} - 127997735782550979360n^{138} + 201365232058175493648n^{135} - 249720798515946840048n^{132} + 359479190251396313824n^{129} - 387730240615006248352n^{126} + 546306682402072020354n^{123} - 536403309612829825902n^{120} + 709693765215932163384n^{117} - 638436788746011996456n^{114} + 766739177521304686844n^{111} - 678807033208335461060n^{108} + 728604118638754426488n^{105} - 638436788746011996456n^{102} + 566425502212037094114n^{99} - 536403309612829825902n^{96} + 382663780961346390496n^{93} - 387730240615006248352n^{90} + 204054671769085217808n^{87} - 249720798515946840048n^{84} + 88855436438949670752n^{81} - 127997735782550979360n^{78} + 25787085845190439324n^{75} - 57123279707358859780n^{72} + 4007373039681507312n^{69} - 18831830242753062096n^{66} - 522013193809718760n^{63} - 4471733528882428008n^{60} - 261725392688093200n^{57} - 929843481131980112n^{54} - 25150781655318804n^{51} - 124533892690899060n^{48} + 7736255993505696n^{45} - 9326892512301792n^{42} + 329330091482992n^{39} - 282571560087184n^{36} - 52628065523424n^{33} - 21640544005728n^{30} - 3423412457169n^{27} + 395751972327n^{24} - 45023273708n^{21} + 3320546180n^{18} + 301985898n^{15} + 52083786n^{12} - 567756n^{9} + 35268n^{6} + 343n^{3} - 1$
$7$ $1$ $\tiny n^{223} + 73n^{221} - 2738n^{219} - 21n^{218} - 11867n^{217} - 504n^{216} - 442788n^{215} - 18648n^{214} - 2395824n^{213} + 154548n^{212} + 45973674n^{211} - 2056104n^{210} - 216916410n^{209} + 2487672n^{208} - 858370548n^{207} - 49704750n^{206} - 6019536950n^{205} + 1293790968n^{204} - 6272422052n^{203} + 3527532504n^{202} - 145611707840n^{201} - 19732629900n^{200} + 162842491357n^{199} + 16839964584n^{198} + 2643736757889n^{197} + 651720025224n^{196} - 9914726637690n^{195} + 797832288243n^{194} + 11872801130937n^{193} + 4778688863808n^{192} + 54682056828768n^{191} - 2752816091328n^{190} + 42921576642432n^{189} + 74977263549216n^{188} + 394726829106352n^{187} + 64221059024064n^{186} + 821578751030800n^{185} - 303991397444160n^{184} + 519123919357600n^{183} + 1988675312056368n^{182} + 8044005325792048n^{181} - 1897953932531520n^{180} - 6743116002517536n^{179} + 2059260322161600n^{178} - 6291775461022464n^{177} + 24740805794281632n^{176} + 136052941835052180n^{175} - 18813168585109440n^{174} - 143192020200233964n^{173} + 9710680708696896n^{172} - 38631641532032616n^{171} + 75668373814702716n^{170} + 977386584837020324n^{169} - 50538549931017120n^{168} - 665538150245382448n^{167} - 115734784323384864n^{166} + 145666222229739200n^{165} - 452793124281273360n^{164} + 5187787362534054488n^{163} + 806174413817128992n^{162} - 1832521178744008152n^{161} - 383855705045949024n^{160} + 605591246599920336n^{159} - 6134487254339699592n^{158} + 19662682113001872216n^{157} + 3925739658222262944n^{156} - 730121067989970096n^{155} + 2464627951903666464n^{154} - 6597848322682530816n^{153} - 36163416643466833296n^{152} + 63341005963509292852n^{151} + 11990850333587882208n^{150} + 17753389806229343140n^{149} + 12242052333943981920n^{148} - 48194702095038787304n^{147} - 128723972783899792980n^{146} + 161946271591214306020n^{145} + 23503703987230614720n^{144} + 71123985985434998304n^{143} + 31504061466236182464n^{142} - 160837630354083921792n^{141} - 338434929802696627872n^{140} + 337243973886317363280n^{139} + 38269257946749799488n^{138} + 188096429130302154864n^{137} + 37642700187699087168n^{136} - 402551056817488319136n^{135} - 680060137129160120496n^{134} + 602436271369366611664n^{133} + 36513480196637752896n^{132} + 365709594686500548256n^{131} + 40497649896824020800n^{130} - 741883938124821325568n^{129} - 1159459265836532430624n^{128} + 915951226723778164942n^{127} + 1778434526383180992n^{126} + 572457630904096994142n^{125} - 4719988173476155968n^{124} - 1156542755772156111804n^{123} - 1632227423105459205126n^{122} + 1217497632360789320838n^{121} - 30987992507585061456n^{120} + 738727103925149091816n^{119} - 45193028654183540112n^{118} - 1452026414045668543008n^{117} - 2027256772469246033544n^{116} + 1393585875735484804796n^{115} - 81985600677340343664n^{114} + 816266516307581035556n^{113} - 77692815594779404464n^{112} - 1580184791895221067832n^{111} - 2149477999531241003412n^{110} + 1400717434571971715324n^{109} - 77692815594779404464n^{108} + 738802367871115614312n^{107} - 81985600677340343664n^{106} - 1424934709158290199936n^{105} - 2027256772469246033544n^{104} + 1220403837909074978502n^{103} - 45193028654183540112n^{102} + 582598436017266814110n^{101} - 30987992507585061456n^{100} - 1108704188606423962764n^{99} - 1632227423105459205126n^{98} + 917124559704170953678n^{97} - 4719988173476155968n^{96} + 365154387831184118944n^{95} + 1778434526383180992n^{94} - 718104062520679564160n^{93} - 1159459265836532430624n^{92} + 604543454626370963920n^{91} + 40497649896824020800n^{90} + 185622229446313610352n^{89} + 36513480196637752896n^{88} - 389697376624984233888n^{87} - 680060137129160120496n^{86} + 333022937924647981392n^{85} + 37642700187699087168n^{84} + 70658789976008488992n^{83} + 38269257946749799488n^{82} - 174375915860388970752n^{81} - 338434929802696627872n^{80} + 160358130606088438756n^{79} + 31504061466236182464n^{78} + 14638045318878429220n^{77} + 23503703987230614720n^{76} - 61005337725569110280n^{75} - 128723972783899792980n^{74} + 63502390865087990452n^{73} + 12242052333943981920n^{72} - 428740817482546608n^{71} + 11990850333587882208n^{70} - 17024945646362445120n^{69} - 36163416643466833296n^{68} + 20632053767828424408n^{67} + 2464627951903666464n^{66} - 927418332041560536n^{65} + 3925739658222262944n^{64} - 2690690670866850864n^{63} - 6134487254339699592n^{62} + 5359247757268077272n^{61} - 383855705045949024n^{60} - 476233125192896560n^{59} + 806174413817128992n^{58} - 94106472416578048n^{57} - 452793124281273360n^{56} + 971609527121070884n^{55} - 115734784323384864n^{54} - 34972974705873516n^{53} - 50538549931017120n^{52} + 14585264272562232n^{51} + 75668373814702716n^{50} + 111172438401799572n^{49} + 9710680708696896n^{48} + 2596839422394336n^{47} - 18813168585109440n^{46} - 2903755235138688n^{45} + 24740805794281632n^{44} + 11079692550353968n^{43} + 2059260322161600n^{42} - 489084178896368n^{41} - 1897953932531520n^{40} - 969026021118560n^{39} + 1988675312056368n^{38} + 439509101205424n^{37} - 303991397444160n^{36} + 69129669094752n^{35} + 64221059024064n^{34} - 92525607640320n^{33} + 74977263549216n^{32} - 770305614663n^{31} - 2752816091328n^{30} + 512992426017n^{29} + 4778688863808n^{28} - 994274068002n^{27} + 797832288243n^{26} + 207681353341n^{25} + 651720025224n^{24} + 151030554268n^{23} + 16839964584n^{22} - 25253990576n^{21} - 19732629900n^{20} + 7740263530n^{19} + 3527532504n^{18} - 795879354n^{17} + 1293790968n^{16} - 161623668n^{15} - 49704750n^{14} - 2367990n^{13} + 2487672n^{12} - 5569764n^{11} - 2056104n^{10} - 376512n^{9} + 154548n^{8} + 30469n^{7} - 18648n^{6} + 937n^{5} - 504n^{4} - 74n^{3} - 21n^{2} + n$

Code used:

def E(A, field=QQ):
    return EllipticCurve(field, [-3*A^4, -A^10 - A^2])

R0.<n> = PolynomialRing(ZZ) F = FractionField(R0) A = n^3 E3 = E(A, field=F) G = E3.point([A^4 + A^2 + 1, A^6 + A^4 + A^2 + 1]) H = E3.point([n^2*(n^8 + 1), 0])

for (k, j) in cartesian_product([[1..7], [0, 1]]): a = n^3 x = (kG + jH)[0] u = a^3 - (a^9 - a)/(x + a^6) b = u v = sqrt( (a^3 - b)(b^3 - a) ) # print(f"a = {a}\nb = {b}\nv = {v}") p, q, r, s = a^3 - b, av, b(a^3 - b), v den = lcm([ expr.denominator() for expr in (p, q, r, s)]) p, q, r, s = pden, qden, rden, s*den div = gcd([p, q, r, s]) p, q, r, s = R0(p/div), R0(q/div), R0(r/div), R0(s/div) # print(f"Check: bool((p+q)^4 + (r-s)^4 == (r+s)^4 + (p-q)^4)}")

f = r - s
f = R0(f / f.content() )
ff = f.homogenize()
fcheck = bool(ff(n, 1)^4 + ff(1, -n)^4 == ff(n, -1)^4 + ff(1, n)^4)

g = R0( (b/n).numerator() )
gcheck = bool(b == n * g(n) / g(1/n) / n^g.degree())

showf, showg = 1, 0
decoration = lambda f: r'\tiny ' if f.degree() &gt; 100 else r'\small ' if f.degree() &gt; 50 else '' 
if fcheck and showf:
    # print(f&quot;f = {f}\n\n&quot;)
    print(f&quot;| <span class="math-container">${k}$</span> | <span class="math-container">${j}$</span> | <span class="math-container">${decoration(f)}{latex(f)}$</span> |&quot;)

if gcheck and showg:
    # print(f&quot;g = {g}\n\n&quot;)
    print(f&quot;| <span class="math-container">${k}$</span> | <span class="math-container">${j}$</span> | <span class="math-container">${decoration(g)}{latex(g)}$</span> |&quot;)


... to be continued ...

dan_fulea
  • 32,856
  • Note there is a hidden symmetry in the computations. Above, in the case $(k,j)=(1,1)$ we obtain the polynomial $$f=(1,1)=f_{G(n^3)+H(n^3)}=n^{7} - 8n^{6} - 17n^{5} - 2n^{4} - 17n^{3} - 20n^{2} + n - 2\ .$$But on the other side, the polynomial from the case $A=n$, $f_1=f_{G(n)}$ was computed having the above "in between", by using after that the Möbius transformation. – dan_fulea Oct 07 '23 at 15:29
  • Thanks so much for this! However, some needed info. For the first family with $A = n$, I did the Mobius transformation to reduce the size of the coefficients, and you can notice that your deg $7, 13, 25, 37$ in Part 1 are the same as mine. For the second family $A = n^3$ in Part 2, I'm not exactly sure what you did, but your deg-$19$ $f_k$ has different coefficients than mine. But it will reduce in size and be the same as mine if you let $n = \frac{v+1}{v-1}$. (I tested it.) Kindly test it on all $f_k$ of Part 2 if the coefficients will reduce in size. – Tito Piezas III Oct 07 '23 at 16:18
  • In this second part, i tried to obtain solutions for $v^2 =(A^3-u)(u-A^3)$ by using $A=n^3$ and the known points $G(n^3), H(n^3)$, and linear combinations of them. Fix such a linear combination. Then associate $p,q,r,s$. It turns out that $r\pm s$ is the "right candidate", since $p\pm q$ drops (expected) degree. OK, then i took $r-s$ when the degree is smaller, and did "the same", exactly as with $p-q$ in the first case. The problem has many hidden symmetries, that i tried to exploit, but so far something in the essence escapes my observations. For this $r-s$ i did the Möbius transformation... – dan_fulea Oct 07 '23 at 16:48
  • I will try to adjust so that the same "known" polynomial appears on the place of $f_{(2,1)}$... – dan_fulea Oct 07 '23 at 16:50
  • I thought so! The Mobius transformation is not needed for the case $A = n^3$. Kindly modify your code to undo it, or some variations thereof, and I'm sure the coefficients of ALL your $f_k$ in Part 2 will reduce in size. I know it will reduce for the deg-$7$ and deg-$19$. – Tito Piezas III Oct 07 '23 at 16:53
  • I noticed that the $f_k$ with $j=0$ have variables that are ALL cubes. They have deg $21, 39,$ etc and are just the deg $7, 13,$ etc in disguise. Maybe just delete them to save space? – Tito Piezas III Oct 08 '23 at 08:06
  • Yes, i also mentioned this, it is the discussion about having "primitive" $f$-polynomials, and explicitly the remark after the table ... "for instance the line for (k,j)=(2,0) produces no new result..." I can of course get rid of them... However in this situation am a puzzled with the continuation. After a long effort in studying specializations of the curve $E(A)$ there is no result i could mention. Also searching in the literature is not simple. The question is explicitly about finding a rational $b=P/Q$ with $P,Q$ of degrees $\le 24$. All our discussion and results go in other direction. – dan_fulea Oct 08 '23 at 16:40
  • On the other side, there is on the MO-page some highly appreciated answer doing the same as me above... i will say here some more words here, but again, as a digression, then move to the MO-question. – dan_fulea Oct 08 '23 at 16:46
  • Yes, in retrospect, the question seems too limited. For the 1st family, the first five points I found had denominators of deg $0,6,8,18,24$, I assumed I may have missed a point and the focus of the question was to determine if there was another of degree $<24$. But as it became clear there was probably none, pursuing the opposite direction of degree $>24$ proved more fruitful, as with more data, several patterns started to emerge, one of which was why the sum of the coefficients for the 1st (and 2nd) family is $2^k$, a question posited in the MO post. – Tito Piezas III Oct 08 '23 at 17:58
2

Part III

... continued answer ...

The following seems to go in a completely wrong direction, but i have no algorithm to find rational points on an elliptic curve defined over some function field, in our case $\Bbb Q(A)$, so "experimental" mathematics was my only hope. The search did not lead to an answer, but since the structure is rich and interesting, i would like to share the data collected so far.


Answer to Question B from the first part:

As said, this is only a try, and so far i have no results, no finality in the question. The strategy i was trying is as follows. Let us plot points for specific values of $A$, and among the points let us try to collect "nice" values sharing the same characteristics, then try to interpolate them to solutions. This strategy was motivated by the fact, that we have "high ranks" that should "statistically" not show up in such a generosity.

Coding, we get the following experimental results for some first few special values of $A$: $$ \small \begin{array}{|c|c|c|c|c|} \hline A & \operatorname{rank} E(\Bbb Q) & x\text{-generator(s)} & x\text{-torsion} & A^4 + A^2 + 1\\\hline 2 & 1 & 21 & & 21\\\hline 3 & 1 & 91 & & 91\\\hline 4 & 1 & 273 & & 273\\\hline 5 & 1 & 651 & & 651\\\hline 6 & 1 & 1333 & & 1333\\\hline 7 & 1 & 2451 & & 2451\\\hline 8 & 2 & 2036,\ \frac{8412452}{841} & 1028 & 4161\\\hline 9 & 1 & 6643 & & 6643\\\hline 10 & 1 & 10101 & & 10101\\\hline 11 & 1 & 14763 & & 14763\\\hline 12 & 1 & 20881 & & 20881\\\hline 13 & 2 & \frac{4545243}{361},\ 28731 & & 28731\\\hline 14 & 3 & \frac{399729}{16},\ \frac{6431629}{225},\ 38613 & & 38613\\\hline 15 & 2 & \frac{1931590}{81},\ 50851 & & 50851\\\hline 16 & 1 & 65793 & & 65793\\\hline 17 & 2 & \frac{4475507}{169},\ 83811 & & 83811\\\hline 18 & 1 & 105301 & & 105301\\\hline 19 & 3 & 45794,\ \frac{30280019}{529},\ 130683 & & 130683\\\hline 20 & 2 & \frac{10818628305}{139129},\ 160401 & & 160401\\\hline 21 & 2 & 194923,\ \frac{5836400603754633832651}{13152400818462889} & & 194923\\\hline 22 & 1 & 234741 & & 234741\\\hline 23 & 2 & \frac{114559169060883}{1610657689},\ 280371 & & 280371\\\hline 24 & 2 & 332353,\ ? & & 332353\\\hline 25 & 1 & 391251 & & 391251\\\hline 26 & 2 & \frac{7000009057179489}{26274464836},\ 457653 & & 457653\\\hline 27 & 1 & 81171 & 59058 & 532171\\\hline 28 & 2 & \frac{8666934584396704161561}{42588200894689921},\ 615441 & & 615441\\\hline 29 & 2 & \frac{518493849366}{6880129},\ 708123 & & 708123\\\hline 30 & 2 & \frac{342151070596465}{1984524304},\ 810901 & & 810901\\\hline 31 & 2 & 924483,\ \frac{84397882725302259}{46604606161} & & 924483\\\hline 32 & 1 & 1049601 & & 1049601\\\hline 33 & 2 & \frac{1125392563}{3969},\ 1187011 & & 1187011\\\hline 34 & 2 & 1337493,\ \frac{1129196031078915469}{5168415762225} & & 1337493\\\hline 35 & 1 & 1501851 & & 1501851\\\hline 36 & 2 & 1680913,\ ? & & 1680913\\\hline 37 & 3 & 1875531,\ \frac{795903707}{289},\ 5777291 & & 1875531\\\hline 38 & 2 & 2086581,\ ? & & 2086581\\\hline 39 & 2 & \frac{3726408361}{6724},\ 2314963 & & 2314963\\\hline 40 & 2 & 2561601,\ \frac{128378784660}{20449} & & 2561601\\\hline 41 & 2 & \frac{1109533185123}{1042441},\ 2827443 & & 2827443\\\hline 42 & 1 & 3113461 & & 3113461\\\hline 43 & 1 & 3420651 & & 3420651\\\hline 44 & 2 & 3750033,\ ? & & 3750033\\\hline 45 & 2 & \frac{36642355}{49},\ 4102651 & & 4102651\\\hline 46 & 1 & 4479573 & & 4479573\\\hline 47 & 2 & 4881891,\ ? & & 4881891\\\hline 48 & 2 & 5310721,\ \frac{266508056958404120212}{4088528484121} & & 5310721\\\hline 49 & 1 & 5767203 & & 5767203\\\hline 50 & 1 & 6252501 & & 6252501\\\hline 51 & 2 & \frac{102017949331}{67081},\ 6767803 & & 6767803\\\hline 52 & 1 & 7314321 & & 7314321\\\hline 53 & 2 & \frac{1967932718651}{2621161},\ 7893291 & & 7893291\\\hline 54 & 1 & 8505973 & & 8505973\\\hline 55 & 1 & 9153651 & & 9153651\\\hline 56 & 2 & \frac{839981748132}{737881},\ 9837633 & & 9837633\\\hline 57 & 1 & 10559251 & & 10559251\\\hline 58 & 1 & 11319861 & & 11319861\\\hline 59 & 1 & 12120843 & & 12120843\\\hline 60 & 1 & 12963601 & & 12963601\\\hline 61 & 1 & 13849563 & & 13849563\\\hline 62 & 1 & 14780181 & & 14780181\\\hline 63 & 1 & 15756931 & & 15756931\\\hline 64 & 2 & 1258256,\ \frac{2530745379542612466391184}{82113604688731225} & 1048592 & 16781313\\\hline 65 & 1 & 17854851 & & 17854851\\\hline 66 & 1 & 18979093 & & 18979093\\\hline 67 & 2 & 20155611,\ ? & & 20155611\\\hline 68 & 1 & 21386001 & & 21386001\\\hline 69 & 2 & 22671883,\ ? & & 22671883\\\hline 70 & 1 & 24014901 & & 24014901\\\hline 71 & 2 & 25416723,\ \frac{88630245334868330091439139}{2375291118001806601} & & 25416723\\\hline 72 & 1 & 26879041 & & 26879041\\\hline 73 & 1 & 28403571 & & 28403571\\\hline 74 & 2 & 29992053,\ ? & & 29992053\\\hline 75 & 2 & \frac{24064744265813313544825}{1336093444022544},\ 31646251 & & 31646251\\\hline 76 & \text{in }[1, 3] & 33367953,\ ? & & 33367953\\\hline 77 & 2 & \frac{8688056181995742171}{3698509768801},\ 35158971 & & 35158971\\\hline 78 & 2 & \frac{332795468793316812570193}{149256682450450576},\ 37021141 & & 37021141\\\hline 79 & 1 & 38956323 & & 38956323\\\hline 80 & 2 & \frac{61775559502580}{3433609},\ 40966401 & & 40966401\\\hline 81 & 2 & 43053283,\ ? & & 43053283\\\hline 82 & 2 & 45218901,\ ? & & 45218901\\\hline 83 & 2 & 47465211,\ \frac{1476196378725808884495051}{12882890828969329} & & 47465211\\\hline 84 & 3 & \frac{49900033}{16},\ \frac{74210525761}{10000},\ 49794193 & & 49794193\\\hline 85 & 2 & 52207851,\ \frac{97984856275}{9} & & 52207851\\\hline 86 & 1 & 54708213 & & 54708213\\\hline 87 & 1 & 57297331 & & 57297331\\\hline 88 & 1 & 59977281 & & 59977281\\\hline \end{array} $$

Used code:

import warnings
warnings.filterwarnings("ignore", category=DeprecationWarning)

def E(A, field=QQ): return EllipticCurve(field, [-3*A^4, -A^10 - A^2])

for A in [2..130]: EA, ok = E(A), True try: rA = EA.rank(algorithm='pari', pari_effort=6) # or use pari_effort=... gens = EA.gens() except RuntimeError: ok = False

if not ok:
    rminA, rmaxA, gens = EA.simon_two_descent()
    rA = rminA if rminA == rmaxA else f'\\text{{in }}[{rminA}, {rmaxA}]'
    gens += [('?', '?', '?')]

xgens = [G[0] for G in gens]
xtors = [P[0] for P in EA.torsion_points() if P]
xgens_info = ',\\ '.join([latex(x) for x in xgens])
xtors_info = ',\\ '.join([latex(x) for x in xtors])
print(f'{A} &amp; {rA} &amp; {xgens_info} &amp; {xtors_info} &amp; {A^4+A^2+1}\\\\\\hline')

For $A=24,36,38,47,\dots$ we have some problems with the curve, exact information is expensive, i did not want to wait or invest supplementary effort. So i considered the cases in detail, and added the information i could find in a cheap manner.

At any rate, this is an interesting family! Yes, we have a point, $G=G(A)$, on the curve, but very often the rank jumps, is two or even three.

Can we guess a formula that produces for special values of $A$ some of these points (that avoid the last column)?

We have rank two in very many cases, slightly unexpected. We also have rank three in many cases, sometimes also a further integral point. Let us look closer for such cases.

for A in [2..1000]:
    EA, special_case = E(A), False
    try:
        rA = EA.rank(algorithm='pari', pari_effort=6) 
    except RuntimeError:
        continue
    if rA >= 3:
        special_case = True
    rminA, rmaxA, gens = EA.simon_two_descent()
    gens = [gen for gen in gens if gen[0] != A^4 + A^2 + 1]
    if not gens:    continue
    for gen in gens:
        if gen[0] in ZZ:    # integer x-component of the generator gen
            special_case = True
if not special_case:
    continue
xgens = [G[0] for G in gens]
xtors = [P[0] for P in EA.torsion_points() if P]
xgens_info = ',\\ '.join([latex(x) for x in xgens])
xtors_info = ',\\ '.join([latex(x) for x in xtors])
print(f'{A} &amp; {rA} &amp; {xgens_info} &amp; {xtors_info} &amp; {A^4+A^2+1}\\\\\\hline')

This produces "interesting" cases of rank at least three, or of further integral points (not equal $A^4+A^2+1$).

$$ \small \begin{array}{|c|c|c|c|c|} \hline A & \operatorname{rank} E(\Bbb Q) & x\text{-generator(s) not in last column} & x\text{-torsion} & A^4 + A^2 + 1\\\hline 8 & 2 & 2036,\ \frac{794809}{576} & 1028 & 4161\\\hline 14 & 3 & \frac{399729}{16} & & 38613\\\hline 19 & 3 & 45794,\ \frac{30280019}{529} & & 130683\\\hline 27 & 1 & 81171 & 59058 & 532171\\\hline 37 & 3 & 5777291,\ \frac{795903707}{289} & & 1875531\\\hline 64 & 2 & 1258256 & 1048592 & 16781313\\\hline 84 & 3 & 4911673,\ \frac{18023706577}{4489} & & 49794193\\\hline 91 & 2 & 7197059 & & 68583243\\\hline 96 & 3 & \frac{2647335815716}{18769} & & 84943873\\\hline 120 & 3 & \frac{48127846308905380}{123899161},\ \frac{288296261406356905}{890604649} & & 207374401\\\hline 123 & 3 & \frac{231964338483931}{9541921},\ \frac{417074779969}{14400} & & 228901771\\\hline 125 & 2 & 10986275 & 9765650 & 244156251\\\hline 136 & 3 & \frac{2051573723573658937}{156149054649} & & 342120513\\\hline 203 & 3 & \frac{10407357398154748975563}{130392259237249} & & 1698222891\\\hline 208 & 4 & 55274856,\ 706387656,\ \frac{63230890024}{1089} & & 1871816961\\\hline 235 & 3 & \frac{50518832434}{225},\ \frac{458320713361}{1296} & & 3049855851\\\hline 244 & 3 & \frac{22762837342937}{94249} & & 3544594833\\\hline 253 & 2 & 905019779 & & 4097216091\\\hline 293 & 3 & \frac{26924779971}{121},\ \frac{1080248226102}{6241} & & 7370136651\\\hline 323 & 4 & \frac{979713058691}{1369},\ \frac{22452449455392798614}{11223919249},\ \frac{24533523939}{49} & & 10884644571\\\hline 336 & 3 & \frac{13970368017736}{11449} & & 12745619713\\\hline 343 & 2 & 300129851,\ 661067379 & 282475298 & 13841404851\\\hline 355 & 3 & \frac{200494304992569}{264196},\ \frac{27386753583865}{86436} & & 15882426651\\\hline 371 & 2 & 3284327859 & & 18945182523\\\hline 397 & 3 & \frac{1240030242867079301243414}{1709268041457025},\ \frac{58047923899592676451469945539}{33111925505392950225} & & 24840754491\\\hline 410 & 3 & \frac{237801960102719541}{391129729} & & 28257778101\\\hline 411 & 2 & 3675864514 & & 28534473163\\\hline 512 & 2 & 1124872256 & 1073741888 & 68719738881\\\hline 596 & 3 & \frac{8295417915248563613121}{545750085001},\ \frac{7282229715761304542294788348161}{1260660010204794559684} & & 126178761873\\\hline 617 & 4 & \frac{3274646434089579}{697225},\ \frac{218087926550930537331}{25615682401} & & 144924495411\\\hline 621 & 3 & \frac{676709709883}{121} & & 148719366523\\\hline 635 & 3 & \frac{83039169145}{36} & & 162590803851\\\hline 665 & 3 & \frac{65247385339776551611078538699}{14967955915872636121} & & 195563392851\\\hline 729 & 1 & 3617538651 & 3486784482 & 282430067923\\\hline 773 & 3 & \frac{152441772401646443}{117649},\ \frac{48373410927366}{3721} & & 357041503371\\\hline 835 & 3 & \frac{17268347636449}{324},\ \frac{13028138691927759517678155}{11628625145929} & & 486123397851\\\hline 870 & 3 & \frac{38382901872902245}{896809},\ \frac{13407528500002562545}{1578631824} & & 572898366901\\\hline 896 & 3 & \frac{3755341305858106529104}{452774223225},\ \frac{17620004032798127988470635298827296}{1830919651481575396646641} & & 644514332673\\\hline 970 & 3 & \frac{64601141702720001}{3625216} & & 885293750901\\\hline 1000 & 1 & 10303030100 & 10000000100 & 1000001000001\\\hline \end{array} $$


To see what i wanted, i will say some more few words. In the above list, the case $A=37$ is, say, interesting, we have a further integer as solution. This is $N=N(A)=3A^4 + 3A^3 + 2A^2+3A$. So how far is this polynomial in $A$ from a family lift? We plug in and factor: $$ x^3 -3A^2x -A^2(A^8+1)\Big|_{\text{computed in }N(A)}= (27a^4 + 27a^3 - a^2 + 29a - 1)(a^2 + 1)^2(a + 1)^2a^2\ . $$ Almost all factors are squares. But one is not. No luck. Here one can search further for cases when the remained factor is a square, again an elliptic curve (with rational point for $a=A=37$), but this is an other story...


A final note on the coding. Well, sage delivers for the $(u,v)$-quartic equation $v^2=(A^3-u)(u^3-A)$ over $\Bbb Q(A)$ in a second the corresponding birational elliptic curve in the form $y^2 = x^3 + fx + g$, with $f,g$ as in the answer. They also provide formulas for $y=Y/Z^3$, $x=X/Z^2$ so that $Y^2=X^3 +fXZ^4+gZ^6$, in terms of the polynomials $X,Y,Z$ in the $(u,v)$ variable.

But the used Möbius transformation (or equivalent) does not use my (human) choice, sage gives the choice making the simpler $Z=-v$.

Here is the code anyways...

R0.<A> = PolynomialRing(QQ)
F = FractionField(R0)
R.<u,v> = PolynomialRing(F)
pol = -v^2 + (A^3 - u)*(u^3 - A)

f, g = WeierstrassForm(pol) X, Y, Z = WeierstrassForm(pol, transformation=True)

u0 = A * (4 + A^2 + 10A^4 + A^6) / (1 + 10A^2 + A^4 + 4A^6) v0 = 2 (A - 1) * A * (A + 1)
* (A^2 - 4A - 1) (A^2 + 4A - 1) (A^2 + 1)^2 * (A^4 + 1)
/ (4A^6 + A^4 + 10A^2 + 1)^2

X0, Y0, Z0 = X.subs(u=u0), Y.subs(u=u0), Z.subs(v=v0) x0, y0 = X0/Z0^2, Y0/Z0^3

E = EllipticCurve(F, [f, g]) P0 = E.point([x0, y0])

G = E.point([A^4 + A^2 + 1, A^6 + A^4 + A^2 +1])

... and so on ...

dan_fulea
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