1. find all natural numbers $a, b$ and $n$ so that the following equation holds:

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

## Solution:
The equation is symmetrical in $a$ and $b$, therefore we can assume oBdA $a \leq b$. If $a \geq 3$, then the LS of the equation is divisible by 3, but the RS is not, contradiction. If $a=1$, then $1+b!=2^{n}$ follows. For $b \geq 2$, the LS is odd and greater than 1, contradiction. Consequently, $b=1$ and therefore $n=1$.

If $a=2$, then $2+b!=2^{n}$ follows analogously. For $b \geq 4$ the LS is not divisible by 4 and is greater than 4, contradiction. Consequently, $b=2$ and $n=2$ or $b=3$ and $n=3$.

The only solution triples $(a, b, n)$ are therefore

$$
(1,1,1),(2,2,2),(2,3,3),(3,2,3)
$$