Suppose $p$ is a prime of the form $4k+1$ , then $4|p-1=|\mathbb Z_p^*|$ , as $\mathbb Z^*_p$ is a cyclic group , so there is $\bar x \in \mathbb Z_p^*$ such that $o(\bar x)=4$ , then $o(\bar x^2)=2$ , now in any field of odd characteristic , $-1$ is the only element of multiplicative order $2$ , so $\bar x ^2=-\bar1=\bar{-1}$ , hence $p|x^2+1$ i.e. $-1$ is a quadratic residue of $p$ . Is this proof correct ? Moreover , is there any other proof , except the usual proofs by Wilson's theorem and quadratic reciprocation theory , that $-1$ is a quadratic residue of primes of the form $4k+1$ ? Moreover I am looking for some proofs of the fact $\mathbb Z^*_p$ is cyclic and then in general that any finite subgroup of the multiplicative group of a field is cyclic , can someone please provide some links or references . Please help . Thanks in advance .
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(side-note: by $\mathbb Z_p$, you really mean $\mathbb Z/p\mathbb Z$; the $p$-adic ring $\mathbb Z_p$ is another beast). For the finite-subgroup-is-cyclic, it all goes down to: the equation $x^n=1$ has at most $n$ solutions in a field. Assume that $G \subset k^{\times}$ is not cyclic; it then contains a group $(\mathbb Z/n\mathbb Z)^2$, and therefore $n^2$ solutions to $x^n=1$. Poof, contradiction! – Circonflexe Apr 17 '15 at 11:28
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If you look at Zagier's one-sentence-proof of Fermat's theorem on sums of two squares, you will notice that every prime of the form $p=4k+1$ can be written as the sum of two squares: from $$ a^2+b^2 = p $$ it follows that $-1$ is a quadratic residue $\!\!\pmod{p}$.
Obviously, the last sentence also follows from the fact that the Legendre symbol satisfies: $$\left(\frac{-1}{p}\right)=(-1)^{\frac{p-1}{2}}.$$
Jack D'Aurizio
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1But the thing is mostly actually the other way round , and by the way Zagier's proof is not actually one sentence if written down in "full detail" verifying all the involutions .... – Apr 17 '15 at 10:17