More $a^3+b^3+c^3=(c+1)^3,$ and $\sqrt[3]{\cos\tfrac{2\pi}7}+\sqrt[3]{\cos\tfrac{4\pi}7}+\sqrt[3]{\cos\tfrac{8\pi}7}=\sqrt[3]{\tfrac{5-3\sqrt[3]7}2}$

Question

I. Solutions

In a previous post, On sums of three cubes of form $a^3+b^3+c^3=(c+1)^3$, an example of which is the well-known,

$$3^3+4^3+5^3=6^3$$

we asked if there were polynomial parameterizations higher than 3rd degree. The answer was in the affirmative. In this new question, we ask if more "triplet parameterizations" can be found. These "triplets" will be described below.

First, the equation $a^3+b^3+c^3=(c+1)^3$ can be solved as a quadratic in $c$, namely,

$$c = \frac{-3+\sqrt{12(a^3+b^3)-3}}6$$

Hence the condition is akin to an elliptic curve,

$$P(a,b):=12(a^3+b^3)-3 = y^2$$

but is to be solved by integer polynomials. Ten solutions have been found so far, the first three being,

$$a_1 = n,\quad b_1 = 3n^2+2n+1$$

$$a_{2,3} = 3n^2, \quad b_{2,3} = 6n^2\pm3n+1\quad$$

The next three found by Old Peter and based on,

$$(\color{blue}{9n^3})^3 + (\color{blue}{-9n^3 + 9n^2 - 3n + 1})^3 + (27n^4 - 18n^3 + 9n^2 - 3n)^3 = (27n^4 - 18n^3 + 9n^2 - 3n + 1)^3$$

thus the first triplet,

\begin{align} a_4 = 9n^3,\quad &b_4 = -9n^3 + 9n^2 - 3n + 1\\[6pt] a_5 = \color{blue}{9n^3},\quad &b_5 = 27n^4 + 18n^3 + 9n^2 + 3n + 1\quad\\[6pt] a_6 = \color{blue}{9n^3 + 9n^2 + 3n + 1}, \quad &b_6 = 27n^4 + 18n^3 + 9n^2 + 3n \end{align}

The latter two will yield different $a^3+b^3+c^3=(c+1)^3$ where all terms are positive for $n>0$. For example, let the triplet $a_k,b_k$ as defined and $c_k = \frac16\left(-3+\sqrt{12(a^3+b^3)-3}\right)$, then for $n=1$,

$$9^3 - 2^3 + 15^3 = 16^3\\ 9^3 + 58^3 + 255^3 = 256^3\\ 22^3 + 57^3 + 255^3 = 256^3$$

I don't know why a parameterization for $u_1^3+u_2^3 = u_3^3+u_4^3$ will yield a pair for $v_1^3+v_2^3+v_3^3=v_4^3,$ with all the terms $(u_i, v_i)$ positive for $n>0.$

The next three were found by yours truly and based on,

$$(\color{red}{9n^3 + 6n^2 + 1})^3 + (\color{red}{-9n^3 + 3n^2 - 3n})^3 + (27n^4 + 6n^2)^3 = (27n^4 + 6n^2 + 1)^3$$

thus the second triplet,

\begin{align} a_7 = 9n^3 + 6n^2 + 1,\quad &b_7 = -9n^3 + 3n^2 - 3n\\[6pt] \quad a_8 = \color{red}{9n^3 + 6n^2 + 1},\quad &b_8 = 27n^4 + 18n^3 + 3n^2 + 3n\\[6pt] a_9 = \color{red}{9n^3 + 3n^2 + 3n}, \quad &b_9 = 27n^4 + 18n^3 + 15n^2 + 3n + 1 \end{align}

Again, the latter two will yield all terms positive for $n>0$. For example, let $n=1$,

$$16^3 - 9^3 + 33^3 = 34^3\\ 16^3 + 51^3 + 213^3 = 214^3\\ 15^3 + 64^3 + 297^3 = 298^3$$

And the 10th by Adam Bailey,

$$a_{10} = (3n + 1)(9n^3 + 15n^2 + 10n + 3),\quad b_{10} = (3n + 1)^2(6n^2 + 5n + 3) + 1\qquad $$

---

II. Addendum

There is a curious connection to Ramanujan's work. The two triplets are related by,

$$a_4+b_4 = a_7+b_7 = \color{magenta}{9n^2 - 3n + 1}$$

a polynomial which is quite important to "real cubic fields". The discriminant $d$ of,

$$x^3 + x^2 - (3 n^2 + n)x + n^3=0$$

is $d=-n^2(\color{magenta}{9n^2+3n+1})^2$. Since it is a negative square, the cubic always has three real roots $x_i$. Then we have Ramanujan's nice identity,

$$x_1^{1/3}+x_2^{1/3}+x_3^{1/3} = \sqrt[3]{-(6n+1)+3\sqrt[3]{n(9n^2+3n+1)}}$$

For $n=-1$, we recover his well-known,

$$\sqrt[3]{\cos\tfrac{2\pi}7}+\sqrt[3]{\cos\tfrac{4\pi}7}+\sqrt[3]{\cos\tfrac{8\pi}7}=\sqrt[3]{\tfrac{5-3\,\sqrt[3]7}2}$$

---

III. Questions:

1. Can you find more polynomial solutions to $P(a,b):=12(a^3+b^3)-3 = y^2$, especially if it is a triplet? One may notice they seem to be based on $a^3\color{red}-b^3+c^3 = (c+1)^3$.
2. And why do some solutions manifest as a triplet? (For example, I can't tease out a triplet from Bailey's solution.)

Answer 1

After much effort, we can answer our question. It turns out there are infinitely many polynomial solutions $(a,b,c)$ to,

$$a^3+b^3+c^3 = (c+1)^3$$

hence,

$$12a^3+12b^3-3 = y^2$$

In fact, using our old friend $\color{blue}{k=9m^2-3m+1}$ from the post, we can find four infinite families.

---

I. First and Second.

Let $p = 9m^2,\;q = 9m^2+1,\,$ then,

\begin{align} a_1 &= 9k^2n^3 + 9k p n^2 + 3p^2 n + (54m^4 - 9m^3 + 9m^2 - 3m)\\[6pt] b_1 &= b_2+k\\[6pt] a_2 &= 9k^2n^3 + 9k q n^2 + 3q^2 n + (54m^4 - 9m^3 + 18m^2 +1)\\[6pt] b_2 &= 27k^2 n^4 + 18k(p+q)n^3 + 9(p^2+q^2)n^2 + 3(135m^4 + 18m^3 + 27m^2 + 1)n + (54m^4 + 9m^3 + 9m^2 + 3m) \end{align}

In addition,

$$a_1^3+b_1^3 = a_2^3+b_2^3$$

Numerical example. Let $m=2$ so $k=31$, and $n=0$,

$$822^3 + (978 + 31)^3 + 22968^3 = 22969^3\\ 865^3 \,+ (978 + 0)^3 \,+ 22968^3 = 22969^3$$

---

II. Third and Fourth.

Let $r = -3m,\; s = -3m+1,\,$ then,

\begin{align} a_3 &= 9k^2n^3 + 9k r n^2 + 3r^2 n + (27m^4 - 18m^3 + 9m^2 - 3m)\\[6pt] b_3 &= b_4+k\\[6pt] a_4 &= 9k^2n^3 + 9k s n^2 + 3s^2 n + (27m^4 - 18m^3 + 9m^2 - 3m + 1)\\[6pt] b_4 &= 27k^2n^4 + 18k(r+s)n^3 + 9(r^2+s^2)n^2 + 3(27 m^4 - 18m^3 + 9m^2 - 6m + 1)n - 9m^3 \end{align}

In addition,

$$a_3^3+b_3^3 = a_4^3+b_4^3$$

Numerical example. Let $m=2$ so $k=31$, and $n=1$,

$$7401^3 + (21225 + 31)^3 + 1826583^3 = 1826584^3\\ 7648^3 \,+ (21225 + 0)^3 \,+ 1826583^3 = 1826584^3$$

---

III. Remarks.

Note that $(p,q,r,s)$ are related by,

$$p+s = q+r = \color{blue}{k=9m^2-3m+1}$$

But it seems "triplets" are harder to find, with the one found by Peter the special case $m=0$ of either pair above, and the one I found an isolated case (?).

P.S. A general formula for infinitely other $k$ has been found. See this post's third answer.