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I remember coming across this fact a while ago in a pdf somewhere, but I haven't been able to find it again. Can someone show me how to prove it? I would appreciate easier proofs.

EDIT: I'm very sorry, I forgot to actually state it.

Prove that $-1$ is not a quadratic residue modulo primes of the form $4k+3$.

Thanks!

user26486
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user45220
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    This depends on how much you already know. It follows from the more general statement $a^{\frac{p-1}{2}}\equiv\left(\frac{a}{p}\right)\pmod {p}, \forall a\in\mathbb Z, p$ an odd prime. – user26486 Feb 21 '15 at 16:10
  • Sorry, I haven't done fractions in modular arithmetic yet. Is it possible to explain/refer to where I can learn about them? Thanks! – user45220 Feb 21 '15 at 16:10
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    I was referring to the Legendre symbol. – user26486 Feb 21 '15 at 16:11
  • @user314: Thanks, I'll try to understand it. – user45220 Feb 21 '15 at 16:12
  • Understanding the proof of my statement is not elementary. At least the proof I know requires knowing about primitive roots modulo primes and knowing that $\left(\frac{a}{p}\right)\left(\frac{b}{p}\right)=\left(\frac{ab}{p}\right), \forall$ primes $p$. – user26486 Feb 21 '15 at 16:15
  • @user314: Ok, cheers. Do you recommend any number theory books? I've been looking to buy one for a while. My friend said Burton's Elementary Number Theory is good for beginners. – user45220 Feb 21 '15 at 16:17
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    Sorry, but I don't really know much about English books. I use a book in my own language. I've heard that "Introduction to Number Theory" by Hua Loo Keng is great, but I can't tell you much. If you haven't heard about it, Engel's "Problem Solving Strategies" is very popular among those studying for Math olympiads. – user26486 Feb 21 '15 at 16:21
  • @user314: Thanks a lot, problem solving strategies one sounds good! – user45220 Feb 21 '15 at 16:35

2 Answers2

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$x^2=-1$ holds if and only if $x$ has order $4$. An element of order $4$ exists in $(\mathbb Z/p\mathbb Z)^* \cong C_{p-1}$ if and only if $4|p-1$.

MooS
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  • Thanks, I didn't understand the notation nearing the end of the proof. Also I'm not asking about $x^2=-1$ but $x^2\equiv -1$. (I apologize if I misunderstood anything) – user45220 Feb 21 '15 at 16:11
  • $C_{p-1}$ is the cyclic group with $p-1$ elements. When i write $x^2=-1$, I interpret this as an eqauality in the ring $\mathbb Z/p\mathbb Z$. This the same as the $\equiv$-notation. – MooS Feb 21 '15 at 16:13
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Hint $\ \overbrace{{\rm mod}\ P=4K\!+\!3\!:}^{\large\ \ \ \ P\ -\ 1\,\ =\,\ 2(\color{#c00}{2K+1})}\,\ \ X^{\large 2}\equiv -1\!\!\!\!\!\!\overset{\ \ \ \ \ \ \ (\ \ )^{\Large\color{#c00}{2K+1}}}\Longrightarrow\! X^{\large P\,-\,1}\equiv -1\ $ contra little Fermat

Bill Dubuque
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