Does $(Z/pZ)^×$ contain a primitive $4$-th root of $1$?

Question

Suppose I have a prime number $p$.

Does $( \mathbb{Z}/p \mathbb{Z})^×$ contain a primitive $4$-th root of $1$ ?

Answer 1

For $p$ an odd prime, $\mathbb{Z}/p\mathbb{Z}^*$ will have a primitive root of unity if and only if $p \equiv 1 \bmod 4$. The group $\mathbb{Z}/p\mathbb{Z}^*$ is a cyclic group of order $p-1$. A primitive fourth root of unity will generate a subgroup of $\mathbb{Z}/p\mathbb{Z}^*$ of order $4$, and any subgroup of order $4$ will be generated by a primitive fourth root of unity (subgroups of cyclic groups are cyclic). So we need only to check if $\mathbb{Z}/p\mathbb{Z}^*$ has a subgroup of order $4$. By Lagrange's Theorem, we require $4 \mid p-1$. Since cyclic groups have subgroups of all possible orders, this is sufficient.

Answer 2

@BigMathTimes has given a full answer assuming you know the existence of primitive roots (i.e. the fact that $(\mathbb{Z}/(p))^\times$ is cyclic). To see that $-1$ is a square mod $p$ whenever $p$ is $1$ mod $4$ without using this fact, write $p=4k+1$ and observe that

$$ X^{p-1}-1 = X^{4k}-1 = (X^2-1)(X^2+1)(X^{4k-4}+X^{4k-8}+\dots+1). $$

The left-hand side has $4k$ roots in $\mathbb{Z}/(p)$ by Fermat's little theorem, but the first and third factors on the right-hand side together have at most $4k-2$ roots. Therefore $X^2+1$ must have a root (in fact, two roots); in other words $-1$ is a square. (I learnt this ingenious argument from Imre Leader.)

There's also a proof using Wilson's theorem that pairs up the terms in $(p-1)!$ to see that it's a square in $\mathbb{Z}/(p)$.