Prove that if m is a square integer then m is neither congruent to 2 modulo 5 nor congruent to 3 modulo 5.
I've seen this problem done for modulo 4 and modulo 8 but this doesn't seem to work for me as the 2k doesn't square nicely to a 4. Maybe I'm misunderstanding, but I'm very lost on this problem and where to start/how to prove. Case proof? Proof by contradiction? Any help is appreciated, thanks
Just square the integers mod $5$. We have \begin{align} 0^2 &\equiv 0 \\ 1^2 &\equiv 1 \\ 2^2 &\equiv 4 \\ 3^2 &\equiv 9 \equiv 4 \\ 4^2 &\equiv 16 \equiv 1 \end{align} Note that $2$ and $3$ don't appear on the right hand side of any of the above expressions.