<s>[INST] Solve $x,y\in \mathbb{Z}$ Solve for $x,y\in \mathbb{Z}$
$$x^{6}=y^{2}+53$$
I tried but I couldn't complete 
 [/INST] $$x^6 = y^2 + 53$$
$$x^6-y^2 = 53$$
From $x^6-y^2 = (x^3+y)(x^3-y)$
$$(x^3+y)(x^3-y) = 53$$
The only factors of $53$ are $1$ and $53$ so let: $$x^3-y = 1$$$$x^3+y = 53$$
Adding the $2$ equations we get: $$2x^3 = 54$$ $$x = 3$$$$y = 3^3-1 = 26$$
Our solutions are $x = 3$, $y = 26$.
But notice that in the original equation we are raising $x$ to the $6^\text{th}$ power and $y$ to the $2^\text{nd}$ power so the signs on $x$ and $y$ can be changed so we get $$\boxed{x = \pm3, y = \pm26}$$
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