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I assume that the $$a^6 + 4 b^3 = c^6$$

has no solution in integers. I think this can be solved trivially, but no success so far. I tried to treat this as a $$(a^3)^2 + 4 b^3 = (c^3)^2 \\ (a^2)^3 + 4 b^3 = (c^2)^3$$ But no success.

Could you please help to find any non-trivial integer solution or prove that there is no such.

Gevorg Hmayakyan
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    Try a^6-c^6? That might work – QuIcKmAtHs Jan 03 '18 at 01:01
  • I am trying to deal with $(c^3 - a^3) (c^3 + a^3) = (c - a)(c^2+ca+a^2) (c^3 + a^3) = 4 b^3$ and trying to determine if the $b$ is even or odd. My hope is that this should bring us to contradiction. – Gevorg Hmayakyan Jan 03 '18 at 01:07
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    If there is a solution, then $b$ is a multiple of seven. I'm trying a few different reductions, let's see... – Henrique Augusto Souza Jan 03 '18 at 01:11
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    Assume that there is a non-zero solution, then for the smallest solution it is not hard to show that $a,b,c$ are pairwise coprime and $a,c$ are odd. Then $1+4b^3\equiv1\pmod8$ so $b$ is even. But I suspect you may still have a lot of work to do. – David Jan 03 '18 at 01:14
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    @HenriqueAugustoSouza Modular arithmetic by itself will never give you an answer as you can always take $a\equiv c\not\equiv0\pmod m$ and $b\equiv0\pmod m$. – David Jan 03 '18 at 01:15
  • I feel like you want something stronger then non-trivial. $a=1,b=0,c=1$ is a solution and is non-trivial but is also, probably, not what you want. – Daniel Gendin Jan 03 '18 at 01:21
  • Yes. Under non trivial I mean here no one of $a,b,c$ is $0$. – Gevorg Hmayakyan Jan 03 '18 at 01:23
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    @David i see.. maybe by looking at $\mathbb{Z}[\sqrt[3]{4}]$? This equation having a solution in $\mathbb{Z}$ implies that $(x + \sqrt[3]{4}y)(x^2 - \sqrt[3]{4}xy + y^2\sqrt[3]{16})$ is a cube in this ring for some $x, y \in \mathbb{Z}$. – Henrique Augusto Souza Jan 03 '18 at 01:25
  • @HenriqueAugustoSouza Maybe. I don't know anything about algebraic properties of that ring (e.g., does it have unique factorisation), but perhaps you do. – David Jan 03 '18 at 01:30

1 Answers1

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If $b=0$ then the solutions are trivially $(a,b,c)=(a,0,\pm a)$, so we assume $b\neq 0$. This in turn means that $a\neq \pm c$, or else $b=0$.

Let $u=a^2,v=b,w=c^2$. Then the equation becomes $$ u^3+4v^3=w^3 $$ Since $a\neq \pm c$, we have $u-w = a^2-c^2\neq 0$.

We can linear transform the equation into an Elliptic curve and use existing tools to find that there are no rational points. This will in turn prove that there are no integer solutions.

Transformation to Elliptic curve
Since $u-w\neq 0$, we can let $$ \begin{align*} x &= \frac{-12v}{u-w}&&, & y &= \frac{18(u+w)}{u-w} \end{align*} $$ Since $u,v,w$ are integers, $x,y$ are rationals. Then we check that $$ \begin{align*} y^2 - x^3 + 108 &= \frac{432(u^3+4v^3-w^3)}{(u-w)^3} = 0 \end{align*} $$ Hence we obtained an Elliptic curve $E$ over $\mathbb Q$: $$ \begin{align*} E: y^2 = x^3 - 108 \end{align*} $$ which a Mordell curve with $n=-108$. (More generally a cubic curve with a known rational point can always be transformed into Weierstrass Normal form $y^2 = x^3+ax^2+bx+c$, for example as in here. Our known rational point is $(u,v,w)=(1,0,1)$.)

Rational points on $E$
For any integer solution of our original equation, via the transformation we will obtain a rational point $(x,y)$ on this curve as mentioned earlier. Using Sagemath, we check that indeed it has no rational points other than $\mathcal O$, which concludes the proof. (Our rational transformation cannot land on $\mathcal O$.)

Another way is to check that the rank is $0$ using the database in OEIS for $y^2=x^3-n$ and manually checking that the torsion points are empty. Any torsion point of integral $E$ must have $y=0$ or $y^2$ divides $D= -27c^2 = -314928$ by the Nagell-Lutz theorem, which reduces to a finite check of $y$ in the set $$ \{0,1,2,3,4,6,9,12,18,27,36,54,81,108,162,324\} $$ which we can check has none fits into $y^2=x^3-108$.

Yong Hao Ng
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