If $p$ is a prime. Then $x^{2} \equiv-1\ \pmod p$ has a solution if and only if $p\equiv 1\ \pmod 4$. Please explain in the easiest way.
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1Counterexample: 2 – S.C.B. May 11 '16 at 14:30
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I assume you mean odd primes.
If $$p \equiv 1 \pmod 4 \Rightarrow \left(\left(\frac{p-1}{2}\right)!\right)^2 \equiv -1 \pmod p (\because \text{Wilson's Theorem})$$ So we have shown that if $p \equiv 1 \pmod 4$, $x^2 \equiv -1 \pmod p$ exists.
Now let us show that if $x^2 \equiv -1 \pmod p$, then $p \equiv 1 \pmod 4$. If $$x^2 \equiv -1 \pmod p \Rightarrow (x^2)^{\frac{p-1}{2}} \equiv (-1)^{\frac{p-1}{2}} \equiv 1 \pmod p $$ By Fermat's Little Theorem.
Thus we have $\frac{p-1}{2} \equiv 0 \pmod 2$, and $p \equiv 1 \pmod 4$.
S.C.B.
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@Ayushakj That depends on how you define "special case", but $$\left(\left(\frac{p-1}{2}\right)!\right)^2 \equiv (-1)^{\frac{p+1}{2}} \pmod p $$ using $-x \equiv p-x \pmod p$. – S.C.B. May 11 '16 at 14:37
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Here is a simple proof using group theory and the fact that $U(p)$ is cyclic. Assume $p$ is odd.
$x^{2} \equiv-1 \bmod p$ has a solution if and only if $U(p)$ has an element of order $4$.
Since $U(p)$ is a cyclic group, this happens iff $4$ divides the order of $U(p)$, which is $p-1$.
lhf
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