Find all triples of positive integers $(a, b, p)$ with $p$ prime and 
$$ a^{p}=b!+p $$
(Belgium) Answer: $(2,2,2)$ and $(3,4,3)$.
Clearly, $a>1$. We consider three cases. Case 1: We have $ab$ which is also impossible since in this case we have $b! \leqslant a!a > 1$. Case 2: We have $a > p$. In this case $b! = a^{p} - p > p^{p} - p \geqslant p!$ so $b > p$ which means that $a^{p} = b! + p$ is divisible by $p$. Hence, $a$ is divisible by $p$ and $b! = a^{p} - p$ is not divisible by $p^{2}$. This means that $b(2 p - 1)! + p \geqslant b! + p$. Comment. The inequality $p^{2 p} > (2 p - 1)! + p$ can be shown e.g. by using 
$$ (2 p - 1)! = [1 \cdot (2 p - 1)] \cdot [2 \cdot (2 p - 2)] \cdots \cdots [(p - 1)(p + 1)] \cdot pp$$ ! and so $b \geqslant p + 1$ which implies that $v_{2}((p + 1)!) \leqslant v_{2}(b!) = v_{2}\left(p^{p-1} - 1\right) \stackrel{L T E}{=} 2 v_{2}(p - 1) + v_{2}(p + 1) - 1 = v_{2}\left(\frac{p - 1}{2} \cdot (p - 1) \cdot (p + 1)\right)$, where in the middle we used lifting-the-exponent lemma. On the RHS we have three factors of $(p + 1)!$. But, due to $p + 1 \geqslant 8$, there are at least 4 even numbers among $1, 2, \ldots, p + 1$, so this case is not possible.