Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $p$ is an odd prime. Show that $x^4 \equiv-1$ (mod $p$) has a solution if and only if $p \equiv1$ (mod $8$).

I have proven one similar result '$x^2 \equiv-1$ (mod $p$) has a solution if and only if $p \equiv1$ (mod $4$)'. I try to mimick the proof but i fail. The following is my attempt:

Suppose $x^4 \equiv-1$ (mod $p$) has a solution, say $a$. Then $a^4 \equiv -1$ (mod $p$). The congruence implies that $\gcd(a^4,p)=1 \Rightarrow \gcd(a,p)=1$. By Fermat's little theorem, $a^{p-1} \equiv 1$ (mod $p$). Note that $1 \equiv a^{p-1} \equiv a^{4{\frac{p-1}{4}}} \equiv (-1)^{\frac{p-1}{4}}$ (mod $p$).

But I don know whether $\frac{p-1}{4}$ is an integer or not. If it is not an integer, then the congruence does not hold.

Can anyone guide me?

EDIT: Since $a^4 \equiv -1$ (mod $p$), we have $a^8 \equiv 1$ (mod $p$) $\Rightarrow $ $ord_p(a)=8 \Rightarrow 8 | \phi(p) \Rightarrow p \equiv 1$ (mod $p$)

Suppose $p \equiv 1$ (mod $8$). Then there exists a primitive root for $p$, say $r$. Then we have $r^{p-1} \equiv (r^{4})^{\frac{p-1}{4}} \equiv1$ (mod $p$). Hence there is a solution to $x^4 \equiv-1$ (mod $p$).

Is my proof in edit correct?

share|improve this question
You have already proved half of it on your own. You've proved that if $x^2=-1$ has a solution in $(\mathbb{Z}/p\mathbb{Z})^{\times}$ then $\frac{p-1}{4}=2k$, so $p \equiv 1 \pmod{8}$. Now can you construct a solution for $x^4 = -1$ if $p \equiv 1 \pmod{8}$ like you did earlier with $x^2 = -1$? –  some1.new4u Sep 23 '13 at 8:16

1 Answer 1


Note that $x^4\equiv -1\mod p$ has a solution if and only if there is an element $x$ of order $8$ in $(\mathbb{Z}/p\mathbb{Z})^\times$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.