Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 4 / 6E / a / Solution

Suppose $x^2\equiv -1(\mathrm{mod}p)$. Then $p\nmid x$, so Fermat's little theorem gives $x^{p-1}\equiv 1$. But $p\equiv 3(\mathrm{mod}4)$ makes $(p-1)/2$ odd, and henceSince this $p$ is odd, $1\not\equiv -1(\mathrm{mod}p)$, a contradiction. There is no square root of minus one in this case.