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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?

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  • $\begingroup$ 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 $x^2 = -1$? $\endgroup$ – user66733 Sep 23 '13 at 8:16
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Hint:

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

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By Legendre, since $x^4 \equiv -1 \pmod p$, we have $$\Big(\frac{-1}{p}\Big) = \Big(\frac{x^4}{p}\Big)$$.

Since $p \equiv 1 \pmod 8$, $p = 8k+1$. By Legendre,

$$\Big(\frac{-1}{p}\Big)= (-1)^{(p-1)/2}=(-1)^{((8k+1)-1)/2}= 1$$.

Thus, we have $\Big(\dfrac{x^4}{p}\Big) = x^{4\cdot (p-1)/2} \pmod p=x^{2\cdot(p-1)} \pmod p=1$

Now, by Fermat's Little Theorem, this is clearly a solution.

I'll leave it to you to prove this the other way around, but that should do 'er.

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  • $\begingroup$ Not sure what this is proving. Aren't you using both sides of the equivalence? $\endgroup$ – darij grinberg Sep 18 '15 at 22:42

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