Determine all triples $(a, b, c)$ of positive integers for which

$$
a !+b !=2^{c !}
$$

holds.

(Walther Janous)

Answer. The only solutions are $(1,1,1)$ and $(2,2,2)$.
We can assume without loss of generality that $a \leq b$.

- For $a=b=1$ we get $c=1$, so $(1,1,1)$ is a solution.
- For $a=1$ and $b>1$ the left side is greater than 1 and odd, and therefore cannot be a power of two. Thus, there is no solution in this case.
- For $a=b=2$ we get $c=2$, so $(2,2,2)$ is a solution.
- For $a=2$ and $b=3$ we have $2 !+3 !=8=2^{3}$. However, there is no $c$ such that $c !=3$. Thus, there is no solution in this case.
- For $a=2$ and $b \geq 4$ we have $2 !+b ! \geq 2 !+4 !=26$. Therefore, $c !>4$ must hold. However, the left side is congruent to 2 modulo 4, while the right side is congruent to 0 modulo 4. Thus, there is no solution in this case.
- For $a \geq 3$ and $b \geq 3$ the left side is divisible by 3, while the right side, being a power of two, is not. Thus, there is no solution in this case.

(Reinhard Razen)