Solution
= Solution
Suppose $x^2\equiv-1\pmod p$. Then $p\nmid x$, so <Fermat's little theorem> gives $x^{p-1}\equiv1$. But $p\equiv3\pmod4$ makes $(p-1)/2$ odd, and hence
$$
x^{p-1}=(x^2)^{(p-1)/2}\equiv(-1)^{(p-1)/2}=-1\pmod p.
$$
Since this $p$ is odd, $1\not\equiv-1\pmod p$, a contradiction. \b[There is no square root of minus one in this case.]