Problem 1826

Solution 1. 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 $b < 2p$. 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 $a^{p} \geqslant (p^{2})^{p} > (2p-1)! + p \geqslant b! + p$.

Comment. The inequality $p^{2p} > (2p-1)! + p$ can be shown e.g. by using
$$(2p-1)! = [1 \cdot (2p-1)] \cdot [2 \cdot (2p-2)] \cdots [(p-1)(p+1)] \cdot p < \left(\left(\frac{2p}{2}\right)^{2}\right)^{p-1} \cdot p = p^{2p-1}$$
where the inequality comes from applying AM-GM to each of the terms in square brackets.

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
$$v_{2}((p+1)!) \leqslant v_{2}(b!) = v_{2}\left(p^{p-1} - 1\right) \stackrel{LTE}{=} 2 v_{2}(p-1) + v_{2}(p+1) - 1 = v_{2}\left(\frac{p-1}{2} \cdot (p-1) \cdot (p+1)\right),$$
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.

Solution 2. The cases $a \neq p$ are covered as in solution 1, as are $p = 2, 3$. For $p \geqslant 5$ we have $b! = p(p^{p-1} - 1)$. 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 $q \geqslant 2p-1$, giving
$$b! \geqslant (2p-1)! = [1 \cdot (2p-1)] \cdot [2 \cdot (2p-2)] \cdots [(p-1) \cdot (p+1)] \cdot p > (2p-1)^{p-1} p > p^{p} > p^{p} - p,$$
a contradiction.

Solution 3. The cases $a \neq p$ are covered as in solution 1, as are $p = 2, 3$. Also $b > p$, as $p^{p} > p! + p$ for $p > 2$. The cases $p = 5, 7, 11$ are also checked manually, so assume $p \geqslant 13$. Let $q \mid p+1$ be an odd prime. By LTE
$$v_{q}\left(p^{p} - p\right) = v_{q}\left(\left(p^{2}\right)^{\frac{p-1}{2}} - 1\right) = v_{q}\left(p^{2} - 1\right) + v_{q}\left(\frac{p-1}{2}\right) = v_{q}(p+1)$$
But $b \geqslant p+1$, so then $v_{q}(b!) > v_{q}(p+1)$, since $q < p+1$, a contradiction. This means that $p+1$ has no odd prime divisor, i.e. $p+1 = 2^{k}$ for some $k$.

Now let $q \mid p-1$ be an odd prime. By LTE
$$v_{q}\left(p^{p} - p\right) = 2 v_{q}(p-1)$$
Let $d = v_{q}(p-1)$. Then $p \geqslant 1 + q^{d}$, so
$$v_{q}(b!) \geqslant v_{q}(p!) \geqslant v_{q}\left(q^{d}!\right) > q^{d-1} \geqslant 2d$$
provided $d \geqslant 2$ and $q > 3$, or $d \geqslant 3$.
If $q = 3, d = 2$ and $p \geqslant 13$ then $v_{q}(b!) \geqslant v_{q}(p!) \geqslant v_{q}(13!) = 5 > 2d$. Either way, $d \leqslant 1$.
If $p > 2q + 1$ (so $p > 3q$, as $q \mid p-1$) then
$$v_{q}(b!) \geqslant v_{q}((3q)!) = 3$$
so we must have $q \geqslant \frac{p}{2}$, in other words, $p-1 = 2q$. This implies that $p = 2^{k} - 1$ and $q = 2^{k-1} - 1$ are both prime, but it is not possible to have two consecutive Mersenne primes.

Solution 4. Let $a = p, b > p$ and $p \geqslant 5$ (the remaining cases are dealt with as in solution 3). Modulo $(p+1)^{2}$ it holds that
$$p^{p} - p = (p+1-1)^{p} - p \equiv \binom{p}{1}(p+1)(-1)^{p-1} + (-1)^{p} - p = p(p+1) - 1 - p = p^{2} - 1 \not\equiv 0 \pmod{(p+1)^{2}}$$
Since $p \geqslant 5$, the numbers $2$ and $\frac{p+1}{2}$ are distinct and less than or equal to $p$. Therefore, $p+1 \mid p!$, and so $(p+1)^{2} \mid (p+1)!$.
But $b \geqslant p+1$, so $b! \equiv 0 \not\equiv p^{p} - p \pmod{(p+1)^{2}}$, a contradiction.

Therefore, the only solutions are $(2, 2, 2)$ and $(3, 4, 3)$.