
2. Find all triples $(p, x, y)$ such that $p^{x}=y^{4}+4$, where $p$ is a prime and $x, y$ are natural numbers.


Solution: We begin with the standard factorisation

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

Thus we have $y^{2}-2 y+2=p^{m}$ and $y^{2}+2 y+2=p^{n}$ for some positive integers $m$ and $n$ such that $m+n=x$. Since $y^{2}-2 y+20$, then $p$ divides 4 or $y$. If $p$ divides 4 , then $p=2$. If $p$ divides $y$, then $y^{2}-2 y+2=p^{m}$ shows that $p$ divides 2 and hence $p=2$. But then $2^{x}=y^{4}+4$, which shows that $y$ is even. Taking $y=2 z$, we get $2^{x-2}=4 z^{4}+1$. This implies that $z=0$ and hence $y=0$, which is a contradiction. Thus $m=0$ and $y^{2}-2 y+2=1$. This gives $y=1$ and hence $p=5, x=1$.
