Find all pairs $(x,y)$ of integers such that $$\sqrt[3]{7x^2-13xy+7y^2}=|x-y|+1.$$

Since $x,y$ are symmetric, we can assume $x\geq y$, so the right-hand side is $x-y+1$. If $x=y$, the equation becomes $\sqrt[3]{x^2}=1$, for which $x=\pm 1$.


let $x-y=n\ge0 \implies \sqrt[3]{y^2+ny+7n^2}=n+1 \implies y^2+ny+7n^2-(n+1)^3=0$

$\Delta=4(n+1)^3-27n^2 \implies n=2m \implies (2m+1)^3-27m^2=t^2$

I don't know how to prove the result, but one solution is OK:

$m=\dfrac{k(k+1)}{2} ,(2m+1)^3-27m^2=\dfrac{((k-1)(k+2)(2k+1))^2}{4}$

hope it will give some help.


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