Find All Positive Integer Pairs for (x, y) in sqrt[3]{7x^2-13xy+7y^2}=|x-y|+1

[anemone]

Here is this week's POTW:

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

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[anemone]

No one answered last week's POTW either. However, you can find the official solution below.

Spoiler

WLOG, let $x\ge y$, then we have $7x^2-13xy+7y^2=(x-y+1)^3$.

Now let $x-y=a$ we have

$7a^2+x(x-a)=(a+1)^3 \implies x^2-ax-a^3+4a^2-3a-1=0$.

Now, as $x$ and $y$ are positive integers so the discriminant of the above quadratic in $x$ must be a perfect square.

$4a^3-15a^2+12a+4=(4a+1)(a-2)^2=m^2$ so $4a+1=k^2$ and thus we obtain a family of solution for different values of $k$ for

$x=\dfrac{k^2-1\pm k(k^2-9)}{8}$ and $y=x-\dfrac{k^2-1}{4}=\dfrac{k^2-1\pm k(k^2-9)}{8}-\dfrac{k^2-1}{4}$.