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!
 A: If you are unfamiliar with modular arithmetic, then this might help: If the remainder of $\frac a5$ is $1$, that is another way of saying that there is an integer $r$ such that $a=5r+1$. Squaring both sides, we get $$a^2=(5r+1)^2=25r^2+10r+1=5(5r^2+2r)+1$$which means that the remainder of $\frac{a^2}5$ is also $1$.
Doing the same thing for the other four possibilities will tell you that the remainder is never $3$.
Note that this is exactly the same as the modular arithmetic approach, only phrased in what is hopefully more accessible language.
A: 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 $a^2 \equiv 1 \mod 5$.
A: 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.
