Problem (Number Theory):
Find all triples of positive integers $(a, b, p)$ with $p$ prime and

$$
a^{p}=b !+p
$$

Solution:
Clearly, $a>1$. We consider three cases.

Case 1: We have $a<p$. Then we either have $a \leqslant b$ which implies $a \mid a^{p}-b$ ! $=p$ leading to a contradiction, or $a>b$ which is also impossible since in this case we have $b ! \leqslant a !<a^{p}-p$, where the last inequality is true for any $p>a>1$.

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 $b<2 p$. If $a<p^{2}$ then $a / p<p$ divides both $a^{p}$ and $b$ ! and hence it also divides $p=a^{p}-b$ ! which is impossible. On the other hand, the case $a \geqslant p^{2}$ is also impossible since then $a^{p} \geqslant\left(p^{2}\right)^{p}>(2 p-1) !+p \geqslant b !+p$.

Case 3: We have $a=p$. In this case $b !=p^{p}-p$. One can check that the values $p=2,3$ lead to the claimed solutions and $p=5$ does not lead to a solution. So we now assume that $p \geqslant 7$. We have $b !=p^{p}-p>p !$ 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.

Clearly, $a>1$. We consider three cases.

Case 1: We have $a<p$. Then we either have $a \leqslant b$ which implies $a \mid a^{p}-b$ ! $=p$ leading to a contradiction, or $a>b$ which is also impossible since in this case we have $b ! \leqslant a !<a^{p}-p$, where the last inequality is true for any $p>a>1$.

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 $b<2 p$. If $a<p^{2}$ then $a / p<p$ divides both $a^{p}$ and $b$ ! and hence it also divides $p=a^{p}-b$ ! which is impossible. On the other hand, the case $a \geqslant p^{2}$ is also impossible since then $a^{p} \geqslant\left(p^{2}\right)^{p}>(2 p-1) !+p \geqslant b !+p$.

Case 3: We have $a=p$. In this case $b !=p^{p}-p$. One can check that the values $p=2,3$ lead to the claimed solutions and $p=5$ does not lead to a solution. For $p \geqslant 5$ we have $b !=p\left(p^{p-1}-1\right)$. By Zsigmondy's Theorem there exists some prime $q$ that divides $p^{p-1}-1$ but does not divide $p^{k}-1$ for $k<p-1$. It follows that $\operatorname{ord}_{q}(p)=p-1$, and hence $q \equiv 1$ $\bmod (p-1)$. Note that $p \neq q$. But then we must have $q \geqslant 2 p-1$, giving

$b ! \geqslant(2 p-1) !=[1 \cdot(2 p-1)] \cdot[2 \cdot(2 p-2)] \cdots \cdots[(p-1) \cdot(p+1)] \cdot p>(2 p-1)^{p-1} p>p^{p}>p^{p}-p$, a contradiction.

Final answer: $(2,2,2),(3,4,3)$