Prove that $a^2$ when divided by $5$ cannot have a remainder of $3$. For any natural number $a$, prove that $a^2$ when divided by $5$ cannot have a remainder of $3$. (Hint: What are the possible values of the remainder when $a$ is divided by $5$?
Using the hint, I found that the possible values of the remainder are $1, 2, 3, 4$, and $0$, and $3$ is only a remainder when the last digit of $a$ is $3$ or $8$.
But I'm not sure how to explain this in a proof, and I don't know how to extend it to $a^2$. Any help would be much appreciated!

if $a \equiv 0 \mod 5$, then $a^2 \equiv 0 \mod 5$.
if $a \equiv 1 \mod 5$, then $a^2 \equiv 1 \mod 5$.
if $a \equiv 2 \mod 5$, then $a^2 \equiv 4 \mod 5$.
if $a \equiv -2 \mod 5$, then $a^2  \equiv 4 \mod 5$.
if $a \equiv -1 \mod 5$, then $a^2 \equiv 1 \mod 5$.
