State Fermat's Theorem and Wilson's Theorem.

Let $p$ be a prime.

(a) Show that if $p \equiv 3(\bmod 4)$ then the equation $x^{2} \equiv-1(\bmod p)$ has no solution.

(b) By considering $\left(\frac{p-1}{2}\right)$ !, or otherwise, show that if $p \equiv 1(\bmod 4)$ then the equation $x^{2} \equiv-1(\bmod p)$ does have a solution.

(c) Show that if $p \equiv 2(\bmod 3)$ then the equation $x^{3} \equiv-1(\bmod p)$ has no solution other than $-1(\bmod p)$.

(d) Using the fact that $14^{2} \equiv-3(\bmod 199)$, find a solution of $x^{3} \equiv-1(\bmod 199)$ that is not $-1(\bmod 199)$.

[Hint: how are the complex numbers $\sqrt{-3}$ and $\sqrt[3]{-1}$ related?]