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!

Any integer $a \equiv 0,\pm1,\pm2 \pmod 5$. Since $2^2=4 \equiv -1 \pmod 5$ it follows that $a^2 \equiv 0, \pm 1 \pmod 5$. This proves both that $a^2 \not \equiv 3 \pmod 5$ which was the original question, and also that $a^2 \not \equiv 2 \pmod 5$ for a bonus conclusion.
