What is $x$, if $1! + 2! + 3! + \cdots + (x-1)! + x! = k^2$ and $k$ is an integer?

Question

What is $x$, if $1! + 2! + 3! + \cdots + (x-1)! + x! = k^2$ and $k$ is an integer?

Using trial and error, it is obvious that for $x < 4$, the given equation has solutions only for $x = 1,k = \pm1$ and $x=3, k = \pm 3$. Where to go from here?

Answer 1

It can be proved that for $x \ge 4$, no solutions exist. The expressions

$$\begin{aligned}1! + 2! + 3! + 4! &= 33 \\ 1! + 2! + 3! + 4! + 5! &= 153 \\ 1! + 2! + 3! + 4! + 5! + 6! &= 873 \\ 1! + 2! + 3! + 4! + 5! + 6! + 7! &= 5913\end{aligned}$$ end with the digit $3$. Now, for $x \ge 4$ the last digit of the sum $1! + 2! + 3! + \cdots + x!$ is equal to 3 and therefore, this sum cannot be equal to a square of a whole number $k$ (because the square of a whole number cannot end with $3$).

Therefore, $1$ and $3$ are the only solutions for $x$.