At first I asked this: Proving something with Wilson theorem.
Now I have to prove that if $p=4n+3$ it's impossible to represent $-1$ in the form $x^2$ modulo $p$. How can I prove it?
Thank you!
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Please show us your attempt. – Ahaan S. Rungta Nov 01 '13 at 21:35
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5Also, it would be much better to say what precisely is supposed to be impossible (in this post). – Jakub Konieczny Nov 01 '13 at 21:36
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Do some research on Legendre symbols first. – tc1729 Nov 01 '13 at 21:40
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If you have a square root of $-1$, such an element is of order four in $\Bbb{Z}_p^*$. What does Lagrange say about that? – Jyrki Lahtonen Nov 01 '13 at 21:41
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I get - $p=3+4n=2n-1$. why this impossible (at the contact of Wilson theorem)? I don't understand... Please help me... Thank you! – CS1 Nov 01 '13 at 21:42
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@JyrkiLahtonen - we didn't learn it. we have to use Wilson theorem... – CS1 Nov 01 '13 at 21:43
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2Hint: Use Fermat's little theorem. – Ivan Loh Nov 01 '13 at 21:44
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@IvanLoh, But how? Where Fermat's little theorem can help me here?? Thank you! – CS1 Nov 01 '13 at 21:54
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2You do realise that @Feanor has already posted an answer which uses Fermat's little theorem, right? – Ivan Loh Nov 01 '13 at 21:56
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Right!! Thank you, If still I wont understand something, I'll ask! Thanks!! – CS1 Nov 01 '13 at 22:04
1 Answers
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It is generally true that $a$ is a quadratic remainder (i.e. is of the form $x^2$ for some $x$) if and only if $a^{\frac{p-1}{2}} \equiv 1 \pmod{p}$. Once you know this, just note that for $p \equiv 3 \pmod{4}$ you have $(-1)^{\frac{p-1}{2}} = -1$ and you are done.
Because you only need to show that $-1$ is not a quadratic residue, it will suffice to show one implication (the easy one). So, suppose that $a = x^2$ and $a^{\frac{p-1}{2}} \equiv -1 \pmod{p}$. We get by substituting one equation to another that $x^{p-1} \equiv -1 \pmod{p}$. But this is in contradiction with Fermat's theorem, which finishes the proof.
Jakub Konieczny
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