Prove that the last digit of a square number can only be 0, 1, 4, 5, 6, or 9.
By the division algorithm, we can write every integer as $n = 10q + r$ where $0 \le r < 10$.
We then have
$$ \begin{align} n^2 & = 100q^2 + 20qr + r^2 \\ \\ & \equiv 0 + 0 + r^2 \pmod {10} \\ \\ & \equiv r^2 \pmod{10} \end{align}$$
We consider each case for $r$:
$r=0$ means $r^2 = 0 \equiv 0 \pmod {10}$
$r=1$ means $r^2 = 1 \equiv 1 \pmod {10}$
$r=2$ means $r^2 = 4 \equiv 4 \pmod {10}$
$r=3$ means $r^2 = 9 \equiv 9 \pmod {10}$
$r=4$ means $r^2 = 16 \equiv 6 \pmod {10}$
$r=5$ means $r^2 = 25 \equiv 5 \pmod {10}$
$r=6$ means $r^2 = 36 \equiv 6 \pmod {10}$
$r=7$ means $r^2 = 49 \equiv 9 \pmod {10}$
$r=8$ means $r^2 = 64 \equiv 4 \pmod {10}$
$r=9$ means $r^2 = 81 \equiv 1 \pmod {10}$
So for any integer, the last digit of its square can only be 0, 1, 4, 5, 6, or 9.