Show that if $p$ is an odd prime, with $p = 3 \pmod{4}$, then
$$ (\mathbb{Z}_{p}^{*})^4 = (\mathbb{Z}_{p}^{*})^2 $$
More generally, show that if $n$ is an odd positive integer, where $p = 3 \pmod{4}$ for each prime $p \mid n$, then:
$$ (\mathbb{Z}_{p}^{*})^4 = (\mathbb{Z}_{p}^{*})^2 $$
I was thinking of using Fermats two squares theorem, but I dont really know how to approach the problem.
Any help would be great.
Thanks
Although one can prove it directly as in mercio's answer, it is insightful to deduce it as follows. Lagrange proved (1769) that in a group of odd order $\rm\,2n+1,\:$ every element is a square, since
$$\begin{eqnarray}\rm\, 1 &=&\rm a^{2n+1}\quad\ \text{by Lagrange's Theorem}\\ \rm \Rightarrow\ a &=&\rm a^{2n+2} = (a^{n+1})^2\end{eqnarray}$$
In a field $\rm\,\Bbb F\,$ of order $\rm\, q = 4n+3,\,$ the subgroup $\rm\,S\,$ of squares has odd order $\rm\,(q\!-\!1)/2 = 2n + 1,\,$ hence every square is a square (of a square), $ $ i.e. a fourth power; $ $ i.e. $\rm\ t \in S\:\Rightarrow\: t = s^2,\,\ s\in S,\ $ and, by definition, $\rm\:s \in S\:\Rightarrow\:s = a^2,\ a\in \Bbb F,\:$ hence $\rm\, t = s^2 = (a^2)^2 = a^4.$
Remark $ $ Lagrange's result is a special case of a $\rm\, k$'th power criterion that frequently proves handy. This criterion has a simple conceptual proof (alas, often overlooked). Suppose that we know that $\rm\, g^n = 1,\, $ so exponents on $\rm\,g\,$ can be interpreted $\rm\,mod\ n\!:\, i \equiv m\:$ $\Rightarrow$ $\rm\,g^i = g^{m}.\ $ We, $ $ consequently, $ $
see clearly that $\rm\,g^i\,$ is a $\rm\,k$'th power if $\rm\ mod\ n\!:\, k\mid i,\:$ i.e. $\rm\ i\equiv jk,\,$ so $\rm\,g^i = g^{jk} = (g^j)^k.\,$ By $\rm\color{#C00}{Bezout}$
$$\rm k\,|\, i\:\ (mod\ n)\!\iff\! \exists\,j\!:\ jk\equiv i\:\ (mod\ n)\!\iff\! \exists\, j,m\!:\ jk \!+\! mn = i\color{#C00}{\!\iff\!} (k,n)\,|\, i$$
Hence we have conceptually derived a proof of the following
Theorem $\rm\ \ \ g^n = 1,\,\ (k,n)\mid i\:\Rightarrow\: g^i\,$ is a $\rm\,k$'th power $\ \ $ [Easy $\rm\,k$'th Power Criterion]
Proof $\rm\ \ By\ Bezout,\,\ (k,n)\mid i\:\Rightarrow\:k\mid i\,\ (mod\ n) \Rightarrow\:i\equiv jk\,\ (mod\ n)\Rightarrow\: g^i = g^{jk} = (g^j)^k$
Note $\,\ $ That $\rm\ \ k\,|\, i\:\ (mod\ n)\!\iff\! (k,n)\,|\, i\ \ $ frequently proves conceptually handy, $ $ e.g. see here. $\ $ The reason behind this will become clearer when one studies cyclic groups and (principal) ideals.
Hint : By Fermat's little theorem, $x^{p-1} \equiv 1 \pmod p$. In particular, $x^2 \equiv x^{p+1} = \left(x^\frac{p+1}4\right)^4$
