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Problem: Given: q is an odd squared number - show that: $q\equiv_8 1$

My assumption: $\forall q\in N:\exists a \in Z: a =1\pmod{2}$ and $a^2=q$.

Then I tried to show that it's only true satisfyingly if $\mathrm{gcd}(q,8)\mid 1 \leftrightarrow x\cdot q+y\cdot 8=1$.

But I don't know how to show that it is true for all numbers $q$.

All hints are welcome.

Joffan
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SuperNova
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    Can you rephrase as "Given (some condition), show that (some result)" please, as I can't work out which of your statements are conditions and which results. I could perhaps guess that your meaning is: "Given that q is an odd square number, show that $q\equiv 1 \pmod 8$" - would that be right? – Joffan Jan 22 '15 at 15:31
  • yes that is right. i haved edited the question. – SuperNova Jan 22 '15 at 15:35
  • Does the notation $a\equiv_p b\iff a\equiv b\pmod p$ appear anywhere in scientific writing? – user26486 Jan 22 '15 at 15:43
  • @user314 It certainly has been used on MSE before... off-hand, the only link I can find is here, but that doesn't establish that it is common/accepted. I'm pretty sure I've seen Bill Dubuque use that notation in an answer or two, but I'm not 100% sure. – apnorton Jan 22 '15 at 17:02

2 Answers2

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Any odd number is congruent to $1,3,5,7$ mod $8$. Each of these when squared gives remainder $1$ mod $8$.

voldemort
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$$\begin{align} 1^2 = 1 &\equiv 1 \pmod 8\\ 3^2=9 &\equiv 1 \pmod 8\\ 5^2=25 &\equiv 1 \pmod 8\\ 7^2=49 &\equiv 1 \pmod 8\\ (8q+k)^2 = 8q(8q+k) + 8qk +k^2 &\equiv k^2 \pmod 8\\ \end{align}$$

Joffan
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