## Task B-4.2.

Determine all natural numbers $x, y$ for which $1!+2!+\cdots+x!=y^{2}$.
## Solution.

If $x=1, 1!=1^2$.

If $x=2, 1!+2!=3 \neq y^2$.

If $x=3, 1!+2!+3!=3^2$.

If $x=4, 1!+2!+3!+4!=33 \neq y^2$.

If $x \geqslant 5, x$! ends with the digit 0 and the sum $1!+2!+3!+4!+\cdots+x$! ends with the digit 3.

Since the square of a natural (integer) number $y$ cannot end with the digit 3, the only solutions are

$$
x=1, y=1 \text{ and } x=3, y=3 \text{, i.e., } (x, y) \in\{(1,1),(3,3)\} \text{. }
$$

Note: If students only guess the ordered pairs that are solutions, without verification, assign one point. For the first two points, verification must be done for all from $x=1$ to $x=4$.