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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

user65909
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1 Answers1

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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.

Charles Hudgins
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  • What is the logical/mathematical justification for this? I'm trying to do this in a proof and though I understand that this works I don't know how to get to this – user65909 Feb 05 '20 at 08:10
  • Have you seen that $\mathbb{Z}/5\mathbb{Z}$ has well-defined multiplication and addition? If this notation is unfamiliar to you, have you seen that modular arithmetic has well-defined multiplication and addition? – Charles Hudgins Feb 05 '20 at 08:12
  • @user65909 $\bmod 5!:,$ every integer $n$ is congruent to its remainder $,\bar n := n\bmod 5,,$ where $,0\le \bar n \le 4,,$ thus $,n\equiv \bar n,\Rightarrow, n^2\equiv \bar n^2,$ by the Congruence Power Rule. That's a rigorous justification using congruences. – Bill Dubuque Feb 05 '20 at 15:29