Prove that $ \forall m \in \mathbb{Z}$, $m^2 \pmod 7$ is $0, 1, 2$ or $4$.

Question

Problem : Prove that $ \forall m \in \mathbb{Z}$, $m^2 \pmod 7$ is $0, 1, 2$ or $4$.

Hi all,

I've tried to do this in an odd-even method but it just doesn't seem to work out for me. The only one I can prove is for '$0$' - by taking $m = 0$ or multiples of $7$.

I tried proving by cases but I don't think that would be the ideal proof.

Can someone please help me with what I should do next?

Answer 1

You just have to emurate number of the form of $7k+r$ where $r \in \{ -3,-2,-1,0,1,2,3\}$.

That is is suffices to compute $r^2 \pmod{7}$ where $r \in \{ -3,-2,-1,0,1,2,3\}$.