[ equations in integers ]
[Residues arithmetic (other).]

## [ Products and factorials ]

Solve the equation $1!+2!+\ldots+n!=m^{2}$ in natural numbers.
Consider the last digits of the specified numbers.

## Solution

$n!$ is divisible by 10 for $n \geq 5$. Therefore, for $n \geq 4$, the left side is congruent to $1+2+6+24$, which is 3 modulo 10. A square cannot end in a 3.

The cases $n=1,2,3$ are checked individually.

## Answer

$(1,1)$ and $(3,3)$.