Math Help - Last digit of perfect squre

1. Last digit of perfect squre

Show that the last decimal digit of a perfect square cannot be 2,3,7, or 8.

Any advice? I don't know where to begin to show this...

2. Nevermind... Figured it out..

3. Hello,
Although you figured it out, just look to this proof (if yours is different of course) :

Here is a little table containing the last digit of any number $n$, and the last digit of its square (thanks to Soroban for the array template) :

$\left|\begin{array}{c|cccccccccc}n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ n^2 & 0 & 1 & 4 & 9 & 6 & 5 & 6 & 9 & 4 & 1 \end{array}\right|$

The last digits $2$, $3$, $7$ and $8$ are missing. Therefore, a perfect square cannot have $2$, $3$, $7$ or $8$ as a last digit.