Find all integer solutions to the following equation:

$$
x^{2}+12=y^{4}
$$
Solution. Rearrange the equation and factorize it:

$$
y^{4}-x^{2}=\left(y^{2}-x\right)\left(y^{2}+x\right)=12
$$

The factors of the product can be the following pairs of divisors of 12:

$$
(1 ; 12), \quad(-1 ;-12), \quad(2 ; 6), \quad(-2 ;-6), \quad(3 ; 4), \quad(-3 ;-4)
$$

From the original equation, we can see that $y^{4}$ and $x^{2}$, and thus $y$ and $x$, have the same parity, meaning they are either both even or both odd. Therefore, we only need to consider the following cases: $(2 ; 6)$ and $(-2 ;-6)$.

First, let

$$
\begin{aligned}
& y^{2}-x=2, \quad \text { or } \quad y^{2}-x=6 \\
& y^{2}+x=6, \quad \text { or } \quad y^{2}+x=2
\end{aligned}
$$

From the first system of equations:

$$
2 y^{2}=8, y= \pm 2 \text { and } x=2
$$

From the second system of equations: $2 y^{2}=8, y= \pm 2$ and $x=-2$.

However, if $y^{2}-x=-2$ and $y^{2}+x=-6$, then $2 y^{2}=-8<0$, so the equation has no solution.

Similarly, if $y^{2}-x=-6$ and $y^{2}+x=-2$, then the equation also has no solution.

The complete solutions to the equation are: if $x=2$, then $y=2$ or $y=-2$; or if $x=-2$, then $y=2$ or $y=-2$.