AUT_ABooklet_2023

Answer. The only solutions are $(1,1,1)$ and $(2,2,2)$.

We can assume without loss of generality that $a\le b$.

For $a=b=1$, we get $c=1$, which gives the solution $(1,1,1)$.

For $a=1$ and $b>1$, the left-hand side is bigger than $1$ and odd, therefore, it cannot be a power of $2$ and we do not get a solution in this case.

For $a=b=2$, we get $c=2$, therefore $(2,2,2)$ is a solution.

For $a=2$ and $b=3$, we get $2!+3!=8=2^3$. But there is no $c$ with $c!=3$. Therefore, there is no solution in this case.

For $a=2$ and $b\ge 4$, we get $2!+b!\ge 2!+4!=26$. Therefore, we have $c!>4$. This implies that the left-hand side is congruent to $2$ modulo $4$, while the right-hand side is congruent to $0$ modulo $4$. Therefore, there is no solution in this case.

For $a\ge 3$ and $b\ge 3$, the left-hand side is divisible by $3$ while the power of $2$ on the right-hand side is not. Therefore, there is no solution in this case.

(Reinhard Razen) ☐