Solve $x,y\in \mathbb{Z}$

Question

Solve for $x,y\in \mathbb{Z}$

$$x^{6}=y^{2}+53$$

I tried but I couldn't complete

Answer 1

Hint: You can try write $x^6-y^2=(x^3-y)(x^3+y)$ and use the fact that 53 is a prime number.

Answer 2

$$x^6 = y^2 + 53$$ $$x^6-y^2 = 53$$ 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 $$\boxed{x = \pm3, y = \pm26}$$