Taiwan IMO Selection Camp, 2022

For fixed $n$ there are clearly finitely many solutions; we will show that there is no solution with $n > 100$. So assume $n > 100$. By AM-GM inequality,
$$\begin{aligned} n! &= 2n(n-1)(n-2)(n-3) \cdot (3 \cdot 4 \cdots (n-4)) \\ &\le 2(n-1)^4 \left( \frac{3 + \cdots + (n-4)}{n-6} \right)^{n-6} = 2(n-1)^4 \left( \frac{n-1}{2} \right)^{n-6} < \left( \frac{n-1}{2} \right)^{n-1}, \end{aligned}$$
thus $a, b, c < \frac{n-1}{2}$.

For every prime $p$ and integer $m \ne 0$, let $\nu_p(m)$ denote the $p$-adic valuation of $m$; that is the greatest non-negative integer $k$ for which $p^k$ divides $m$. Legendre's formula states that
$$\nu_p(n!) < \sum_{s=1}^{\infty} \frac{n}{p^s} = \frac{n}{p-1}. \quad (1)$$
If $n$ is odd then $a^{n-1}, b^{n-1}, c^{n-1}$ are squares, and by considering them modulo $4$ we conclude that $a, b, c$ must be even. hence, $2^{n-1} \mid n!$ but that is impossible for odd $n$ because $\nu_2(n!) = \nu_2((n-1)!) < n-1$ by (1). From now on we assume that $n$ is even. If all three numbers $a+b, b+c, c+a$ are powers of $2$ then $a, b, c$ have the same parity. If they are all odd, then $n! = a^{n-1}+b^{n-1}+c^{n-1}$ is also odd, contradicting the assumption that $n$ must be even. If all $a, b, c$ are divisible by $4$, this contradicts $\nu_2(n!) \le n-1$. If, say, $a$ is not divisible by $4$, then $2a = (a+b) + (a+c) - (b+c)$ is not divisible by $8$, and since all $a+b, b+c, c+a$ are powers of $2$ we get that one of these sums equals $4$, so two of the numbers of $a, b, c$ are equal to $2$. Say $a = b = 2$, then $c = 2^r - 2$ and since $c \mid n!$, we must have $c \mid a^{n-1} + b^{n-1} = 2^n$ implying $r = 2$, and so $c = 2$, which is impossible because $n! \equiv 0 \not\equiv 3 \cdot 2^{n-1} \pmod 5$.

So now we assume that the sum of two numbers among $a, b, c$, say $a+b$, is not a power of $2$, so it is divisible by some odd prime $p$. Then $p \le a+b < n$ and so $c^{n-1} = n! - (a^{n-1} + b^{n-1})$ is divisible by $p$. If $p$ divides $a$ and $b$, we get $p^{n-1} \mid n!$,

contradicting (1). Next, using (1) and the Lifting the Exponent Lemma we get
$$\nu_p(1)+\nu_p(2)+\cdots+\nu_p(n) = \nu_p(n!) = \nu_p(n!-c^{n-1}) = \nu_p(a^{n-1}+b^{n-1}) = \nu_p(a+b)+\nu_p(n-1). \tag{2}$$
In view of (2), no number of $1, 2, \dots, n$ can be divisible by $p$, except $a+b$ and $n-1 > a+b$. On the other hand, $p|c$ implies that $p < n/2$ and so there must be at least two such numbers. Hence, there are two multiples of $p$ among $1, 2, \dots, n$, namely $a+b = p$ and $n-1 = 2p$. But this is another contradiction because $n-1$ is odd. This final contradiction shows that there is no solution of the equation for $n > 100$.