2. For which prime numbers $p$ and $q$ is the number $(p+1)^{q}$ a perfect square?
II/2. If $q=2$, then $(p+1)^{2}$ is a perfect square for any prime number $p$. Otherwise, $q$ is odd, so $q=2k+1$ for some natural number $k$. Then $(p+1)^{q}=(p+1)^{2k+1}=(p+1)(p+1)^{2k}$, so the number $(p+1)$ must be a perfect square, that is, $p+1=n^{2}$, from which it follows that $p=(n-1)(n+1)$. Since $p$ is a prime number, this is only possible if $n=2$, in which case $p=3$. Therefore, $4^{q}$ is a perfect square for any prime number $q$.