Fine, I'll give it (Speechless)

Let \(\displaystyle n\) be any integer number. \(\displaystyle n^2\) is its square. Let us look at the last digit of \(\displaystyle n^2\), using arithmetic modulo \(\displaystyle 10\). The following table enumerates all possible cases :

\(\displaystyle \begin{array}{|l|l|}

\hline n \mod 10 &n^2 \mod 10 \\

\hline 0 &0 \\

1 &1 \\

2 &4 \\

3 &9 \\

4 &6 \\

5 &5 \\

6 &6

\end{array}\)

\(\displaystyle \begin{array}{|l|l|}

7 \ \ \ \ \ \ \ \ \ \ \ \ \ &9 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\

8 &4 \\

9 &1 \\

\hline

\end{array}\)

Sorry, I had to cut the table midway, it was too big for MHF.

This shows that a perfect square may only end by the digits \(\displaystyle 0, 1, 4, 5, 6\) and \(\displaystyle 9\). If an integer does not terminate by neither of those digits, it cannot possibly be a perfect square.

\(\displaystyle \boxed{\mathrm{QED}}\)

*This implies that if a number terminates by 3, it cannot be a perfect square.*