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Is there some sophisticated method (or maybe some easy one, though I doubt it) to show that the only solution to $m^2 = 14k^4 - 6k^2 + 1$ in positive integers is $k=1$, $m=3$? Perhaps something around elliptic curves and SAGE computation or whatever seems fine, though the simplest attack will be most appreciated!

When trying this out by hand, the best idea I had what to write it like $m^2 - 14(k^2 - \mbox{something})^2 = \mbox{something}$ and then consider this Pell-type equation, but the presence of $k^2$ in one of the Pell variables makes things much harder.

DesmondMiles
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1 Answers1

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$$m^2=14k^4-6k^2+1\tag{1}$$

An equation $(1)$ is birationally equivalent to the elliptic curve $$Y^2 = X^3 - 68X + 208\tag{2}$$
with $$k = \frac{2X-8}{Y}, m = \frac{X^3-12X^2-144+68X}{Y^2}$$

According to LMFDB , elliptic curve has seven integer points as follows.

$(-6,\pm 20)$, $ \left(4, 0\right) $, $(6,\pm 4)$, $(13,\pm 39).$

Hence we get the positive integer solution $(k,m)=(1,3)$ from $(X,Y)=(6,\pm 4).$

Tomita
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  • I like this approach, but there is one thing I do not understand. You seem to implicitly use that if $(k,m)$ is an integral point on $(1)$ then there is an integral point $(X,Y)$ on $(2)$ such that $$k = \frac{2X-8}{Y}, m = \frac{X^3-12X^2-144+68X}{Y^2}.$$ Can you explain why this is true? Or have I misunderstood? – Servaes Aug 01 '21 at 12:19
  • In order for $(k,m)$ to be an integer, at least $(X,Y)$ must be an integer. – Tomita Aug 01 '21 at 12:32
  • That is exactly my question; why must $X$ and $Y$ be integers? I see that if $(X,Y)$ is a rational point on $(2)$ with $X\neq4$ then $$m=\frac{X^3-12X^2+68X-144}{X^3-68X+208}=\frac{X^2-8X+36}{X^2+4X-52},$$ and hence for $n:=1-m$ we have $$nX^2+4(n-3)X+4(22-13n)=0,$$ which means $X=\tfrac{a}{b}$ for integers $a$ and $b$ with $a\mid4(22-13n)$ and $b\mid n$. But I do not see why $X$ must be integral, i.e. why we can take – Servaes Aug 01 '21 at 13:37
  • In general, since rational points $(X,Y)$ increase rapidly, we can't much expect the integer solution from rational points $(X,Y).$
    For instance, $(X,Y)=(13 , 39 ),(918/49 , 25460/343 ),(34537/6084 , 1052117/474552 ),(28276086/78961 , 150319301164/22188041 ).$
    Substitute $(X,Y)=(\frac{b}{a^2},\frac{c}{a^3})$ to $k=\frac{2X-8}{Y}$, then $k=\frac{2(a-4c^2)c}{b}.$
    Thus, though we may get the integer points $k$ using small rational points $(X,Y)$, we can't get it for this problem.     – Tomita Aug 02 '21 at 02:21
  • BTW, there is a case that integer points of quartic are derived from Mordell-Weil group below.
    It's similar to this problem. (https://math.stackexchange.com/questions/3389407/a-diophantine-equation-related-to-arithmetic-progression-t-n-anbncn/3397172#3397172)     – Tomita Aug 02 '21 at 02:22