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Show that $x^4 \equiv -1 \pmod p $ is solvable $\iff $ $p \equiv 1 \pmod 8$


My attempt : $p$ must satisfy $(-1)^{(p-1)/d}\equiv 1 \pmod p$, where $d = \gcd(4,p-1)$ but I still don't see how this condition implies $p\equiv 1\pmod 8$. Can I get some help thanks!

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  • $\begingroup$ @André Nicolas help $\endgroup$ – pooja Jan 31 '15 at 12:06
  • $\begingroup$ Where $p$ is prime? $\endgroup$ – barak manos Jan 31 '15 at 12:29
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$x^4=-1$ is solvable if and only if the group $(\mathbb Z/p\mathbb Z)^*$ has an element of order $8$. Since $(\mathbb Z/p\mathbb Z)^*$ is a cyclic group of order $p-1$, the assertion follows immediately.

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  • $\begingroup$ Do you think that there is a solution using Bézout relation and without group theory ? I know that Fermat is possible... $\endgroup$ – Maman Jan 21 '16 at 19:51
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For your problem, let $(a,m)=(2,p)$.

$x^{2^a}\equiv -1\pmod{\!m}\iff\text{ord}_m (x)=2^{a+1}$

$(\text{ord}_p (x)=2^{a+1}$ for some $x)\iff (2^{a+1}\mid p-1)$

Proof of '$\Rightarrow$': use $\text{ord}_m (x)\mid h$ with $x^h\equiv 1\pmod{\! m}$ and use little Fermat.

($\text{ord}_m (x)\mid h$ can be proved by contr: $h=\text{ord}_m (x)j+r$ with $0<r<\text{ord}_m (x)$ implies $x^h\equiv x^r\equiv 1\pmod{\!m}$, contradicts $\text{ord}_m (x)$ being the least positive $e$ with $x^e\equiv 1\pmod{\! m}$).

Proof of '$\Leftarrow$': let $p=2^{a+1}k+1$, let $g$ be a primitive root mod $p$.

mod $p$ get $\left(g^k\right)^{2^{a+1}}\equiv 1$ and $\left(g^k\right)^{2^a}\equiv -1$, so $\text{ord}_p (g^k)=2^{a+1}$ by above lemma.

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