1. (4) Solve the equation in natural numbers: $a!+b!=5^{n}$.
1. Solution. Note that if $a \geqslant 2$ and $b \geqslant 2$, then $a! + b!$ is even, and therefore cannot equal $5^{n}$. Let's assume, for definiteness, that $a=1$. We have the equation $1+b!=5^{n}$. Now, if $b \geqslant 5$, then $b!$ is divisible by 5, which means $b!+1$ is not divisible by 5, and again cannot equal $5^{n}$. Therefore, we need to check the cases $b=1, b=2, b=3$, and $b=4$. Only the case $b=4$ works, in which case $n=2$. Thus, we get two solutions: $a=1, b=4, n=2$ and $a=4, b=1, n=2$.