1. In the set of natural numbers, solve the equation

$$
a^{2}+b^{2}+1=c!
$$
Solution. For every integer $x$, it holds that $x \equiv 0,1(\bmod 4)$, so

$$
c!=a^{2}+b^{2}+1 \equiv 1,2,3(\bmod 4)
$$

From this, it follows that $c \leq 3$, because for $c \geq 4$ it holds that $c!\equiv 0(\bmod 4)$. By direct verification for $c=1,2,3$, it is found that the only solutions to the given equation in the set of natural numbers are

$$
(a, b, c)=(1,2,3) \text { and }(a, b, c)=(2,1,3) \text {. }
$$