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.
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.