## Task 20/78

Determine all pairs of natural numbers $(m ; n)$ that satisfy the equation: $\sum_{i=1}^{n} i!=m^{2}$
Due to $\sum_{i=1}^{4} i!=33$ and $i! \equiv 0(\bmod 10)$, if $i>4$, for $n>4$

$$
\sum_{i=1}^{n} i! \equiv 3(\bmod 10)
$$

Since for $m^{2}$ it always holds that $m^{2} \equiv 3(\bmod 10)$, there can be no solutions for $n>4$ (square numbers can only have the units digit 0 or 1 or 4 or 5 or 6 or 9). By trying $n=1 ; 2 ; 3 ; 4$, one obtains the pairs $\left(n_{1} ; m_{1}\right)=(1 ; 1)$ and $\left(n_{2} ; m_{2}\right)=(3 ; 3)$. Apparently, there can be no further solutions.