No square ends in 3

This is an interesting number theory fact that seems strange when taken at face value. Here’s a small proof of it: $$\begin{array}{|c|c|} \hline \text{last digit of number} & \text{last digit of square} \\ \hline 0 & 0 \\ 1 & 1 \\ 2 & 4 \\ 3 & 9 \\ 4 & 6 \\ 5 & 5 \\ 6 & 6 \\ 7 & 9 \\ 8 & 4 \\ 9 & 1 \\ \hline \end{array}$$

It’s easy to see that no square is ending in 3 (or 2, 7 or 8 for that matter). What are the implications of this? Try the following sum:

How many solutions of $x$ exist such that $$\sum_{i=1}^x i! = n^2, \space n \in N$$

In simpler terms, what summations of the factorial give us a perfect square? We can work out the first few by hand $$\begin{align} x = 1 &\implies S_1 = 1! = 1\\ x = 2 &\implies S_2 = S_1 + 2! = 3 \\ x = 3 &\implies S_3 = S_2 + 3! = 9 \\ x = 4 &\implies S_4 = S_3 + 4! = 33 \end{align}$$

For $5!$ and above, the last number is always 0. When we use the above recurrence relation for $x \ge 5$, the last digit will always be 3, and since no square ends in 3, there are only two values of $x$ for which the given equation holds, which are $1$ and $3$. $\blacksquare$