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Are there any coprime integers $x,y$ ( greater than 1 in absolute value) such that $$3y(4x^3-y^3)$$ is a square?

I performed a search on wolfram, I could not find any. Please share if you can find any such pair using a different program or explain why there is none. I would really appreciate it. Thanks.

NumThcurious
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1 Answers1

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You are looking for integer solutions $(x,y,a)$ with $\lvert x\rvert, \lvert y\rvert>1$ and gcd$(x,y)=1$ such that $$3y\cdot (4x^3-y^3)=a^2$$ Notice that this implies \begin{align*}(3y^2+a)^3+(3y^2-a)^3&=54y^6+18a^2y^2\\&=54y^6+18y^2\cdot 3y\cdot (4x^3-y^3)\\&=18y^2\cdot 3y\cdot 4x^3\\&=216\cdot x^3y^3=(6xy)^3\end{align*} Hence, we have concluded that $$(3y^2+a)^3+(3y^2-a)^3=(6xy)^3$$ This is nice, since we can now use Fermat's Last Theorem to infer that one among $3y^2+a, 3y^2-a, 6xy$ is zero. Furthermore $\lvert x\rvert, \lvert y\rvert>1\implies 6xy\not=0$ and, thus, $a=\pm 3y^2$. But it is straightforward to check that this yields $x=y$, which violates gcd$(x,y)=1$. Hence, no solutions.

Dr. Mathva
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  • wow! Very impressive. That's mind boggling that you got to that conclusion( at least for me). I don't understand how you found the implication:$(3y^2+a)^3+(3y^2-a3=(6xy)^3$. Please educate me. – NumThcurious Oct 28 '20 at 14:39
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    @NumThcurious You are not alone. Such a magical solution indeed. That you can complete the cubes like that is pure art. I am also interested in more elaboration of how Dr. Mathva tackled this. :) – Cornman Oct 28 '20 at 14:41
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    I think I saw a similar problem once, but I do not even remember if this was on this site, or maybe in some contest ... anyhow, I remembered one solution that boiled down to find rational points on an elliptic curve, which then simplified to the cubes-equation. I just omitted the part with the elliptic curves, it makes the proof less beautiful :) – Dr. Mathva Oct 28 '20 at 14:53
  • Brilliant! I am so stunned. Thanks. – NumThcurious Oct 28 '20 at 15:38
  • @Dr.Mathva Do you have a reference/recommendation for the method, or on elliptic curves? – Cornman Oct 29 '20 at 07:27