International Mathematical Olympiad

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

Case 1: We have $a<p$. Then we either have $a⩽b$ 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 $b!⩽a!<a^p-p$, where the last inequality is true for any $p>a>1$.

Case 2: We have $a>p$. In this case $b!=a^p-p>p^p-p⩾p!$ 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 $a⩾p^2$ is also impossible since then $a^p⩾(p^2{)}^p>(2p-1)!+p⩾b!+p$.

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 $p⩾7$. We have $b!=p^p-p>p!$ and so $b⩾p+1$ which implies that $$v_2((p+1)!)⩽v_2(b!)=v_2\left(p^{p-1}-1\right)\stackrel{LTE}{=}2v_2(p-1)+v_2(p+1)-1=v_2\left(\frac{p-1}{2}\cdot (p-1)\cdot (p+1)\right),$$ where in the middle we used lifting-the-exponent lemma. On the RHS we have three factors of $(p+1)!$. But, due to $p+1⩾8$, there are at least 4 even numbers among $1,2,\dots ,p+1$, so this case is not possible.

Solution 2. The cases $a\ne p$ are covered as in solution 1, as are $p=2,3$. For $p⩾5$ 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 ${\mathrm{ord}}_q(p)=p-1$, and hence $q\equiv 1\mathrm{mod}(p-1)$. Note that $p\ne q$. But then we must have $q⩾2p-1$, giving $$b!⩾(2p-1)!=[1\cdot (2p-1)]\cdot [2\cdot (2p-2)]\cdots [(p-1)\cdot (p+1)]\cdot p>(2p-1{)}^{p-1}p>p^p>p^p-p,$$ a contradiction.

Solution 3. The cases $a\ne p$ 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 $p⩾13$. 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 $b⩾p+1$, 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 $$v_q\left(p^p-p\right)=2v_q(p-1)$$ Let $d=v_q(p-1)$. Then $p⩾1+q^d$, so $$v_q(b!)⩾v_q(p!)⩾v_q\left(q^d!\right)>q^{d-1}⩾2d$$ provided $d⩾2$ and $q>3$, or $d⩾3$. If $q=3,d=2$ and $p⩾13$ then $v_q(b!)⩾v_q(p!)⩾v_q(13!)=5>2d$. Either way, $d⩽1$. If $p>2q+1$ (so $p>3q$, as $q\mid p-1$) then $$v_q(b!)⩾v_q((3q)!)=3$$ so we must have $q⩾\frac{p}{2}$, 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 $p⩾5$ (the remaining cases are dealt with as in solution 3). Modulo $(p+1{)}^2$ it holds that $$p^p-p=(p+1-1{)}^p-p\equiv (\genfrac{}{}{0ex}{}{p}{1})(p+1)(-1{)}^{p-1}+(-1{)}^p-p=p(p+1)-1-p=p^2-1\not\equiv 0(\mathrm{mod}(p+1{)}^2)$$ Since $p⩾5$, 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 $b⩾p+1$, so $b!\equiv 0\not\equiv p^p-p(\mathrm{mod}(p+1{)}^2)$, a contradiction.

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