What are the fields such that $x^4 = 1$ for every $x$ in the multiplicative group

What are the fields $K$ such that $x^4 = 1$ for every $x \in K^{\ast}$, i.e. such that every element of the multiplicative group is a root of $x^4 - 1$? Of course the finite fields of order $3$ and $5$ are examples, but are this the only ones?

If $K$ is finite, then $K^{\ast}$ is cyclic, hence the equation just holds if $|K^{\ast}| = 4$ or $|K^{\ast}| = 2$, i.e. if for $|K| = p^n$ we have $p^n - 1 = 2$ or $p^n - 1 = 4$, both equations give $p = 3$ or $p = 5$ and $n = 1$. So I guess other field have to be infinite.

• Don't forget the field of two elements. – Justpassingby Dec 23 '15 at 19:54
• Yes, thanks for pointing out! – StefanH Dec 23 '15 at 19:55

Note that the polynomial $x^4-1$ can not have more than $4$ roots, so $|K^*|\le 4$.
Note that a degree $n$ polynomial can have at most $n$ roots over a field. So a field in which all non-zero elements satisfy $x^4=1$ can have at most $4$ non-zero elements, hence at most $5$ elements all together. The only such fields are $\mathbb{F}_2,\mathbb{F}_3$ and $\mathbb{F}_5$, all of which satisfy the condition.
Just adding since no one has said it: Note that, since the group $\mathbb F_q^*$ is cyclic of order $q-1$, the condition that $x^4=1$ for every $x \in \mathbb F_q^*$ is equivalent to $(q-1) \mid 4$. So you simply need to enumerate all prime powers $q$ such that $q-1$ divides into $4$, which doesn't leave very many options.