PROBLEM 4. - Determine all pairs of integers $(x, y)$ that satisfy the equation $x^{2}-y^{4}=2009$.

## Solution:
Given any solution $(x, y)$, it is clear that $(x,-y),(-x, y)$ and $(-x,-y)$ are also solutions, so we can assume without loss of generality that $x, y \geq 0$. Suppose then that this is the case. It is clear that $\left(x-y^{2}\right)\left(x+y^{2}\right)=7^{2} \cdot 41$.

If $x-y^{2}$ and $x+y^{2}$ are not coprime, their greatest common divisor squared divides $2009=7^{2} \cdot 41$, so it is 7 and divides $\left(x+y^{2}\right)+\left(x-y^{2}\right)=2 x$ and $\left(x+y^{2}\right)-\left(x-y^{2}\right)=2 y^{2}$, so there exist non-negative integers $u$ and $v$ such that $x=7 u, \quad y=7 v \quad$ and $\left(u+7 v^{2}\right)\left(u-7 v^{2}\right)=41$. Since both factors must be integers, we have $u+7 v^{2}=41$ and $u-7 v^{2}=1$, so $u=21$ and $v^{2}=\frac{10}{7}$. There are therefore no integer solutions in this case.

If $x-y^{2}$ and $x+y^{2}$ are coprime, one possible case is that $x-y^{2}=1$ and $x+y^{2}=2009$, so $y^{2}=1004$, which is absurd since $31^{2}=961<1004<32^{2}$. The only remaining case is then that $x-y^{2}=41$ and $x+y^{2}=49$, which yields $x=45, y^{2}=4$, so the only solution with non-negative integers is $x=45$ and $y=2$, and the only integer solutions are $(x, y)=( \pm 45, \pm 2)$.