$x^4 = 1$ (mod $p$) implies p = 1 mod 8
1,423
Solution 1
$x^4\equiv 1\pmod{p}\iff \text{ord}_p(x)=8$ (why?).
$x^k\equiv 1\pmod{p}\iff \text{ord}_p(x)\mid k$ (why? hint: for contradiction, let $k=\text{ord}_p(x)h+r$ with $0<r<\text{ord}_p(x)$ and get that $x^r\equiv 1\pmod{p}$, contradiction).
And remember Fermat's Little theorem. This is all you need to know.
Solution 2
By FLT,$$x^{p1}\equiv1\pmod{p}$$
$$x^8\equiv 1\pmod{p}\implies 8\mid p1\implies p\equiv1\pmod{8}$$
The case where $p1 \mid 8$ is impossible because $p\neq 9$ so we need to check only $p=2,3,5$ where,
$$x^4\equiv 1\pmod{2,3,5}$$
so not possible.
(The $x\mid p$ and $p\mid x$ case is easily eliminated so not considered)
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Ludwwwig
Updated on August 01, 2022Comments

Ludwwwig over 1 year
Let $p$ be an odd prime. Show that
$x^4 = 1$ (mod $p$)
has a solution if and only if $\Leftrightarrow p = 1$ (mod $8$)

caffeinemachine almost 8 yearsFor some $x$ or for all $x$?

Ludwwwig almost 8 yearsI have now changed my question; for some x.


Mathmo123 almost 8 yearsYou must be missing some conditions from your argument. $(1)^8 \equiv 1 \pmod 3$ for example, but $3\not\equiv 1\pmod 8$.

cr001 almost 8 yearsYou are right, actually it can be $p1\mid 8$ as well, I will add that case in the proof.

lhf almost 8 yearsThis is one direction only.

Mathmo123 almost 8 years@cr001 in which case, take $1^8 \equiv 1 \pmod 3$. But $(p1)\nmid 1$ and $p\not\equiv 1 \pmod 8$. For your argument to work, you need to know that $x$ has order $8$, which it does in this case. But as written this is wrong.

Ludwwwig almost 8 yearsFor the other direction, let $1 = r^k$, r is a primitive root. Since $r^(k(p1)/4) = 1$ mod 8, we get that (p1)  k/4 (p1). Hence k = 4j. If we let x = r^j we have that x^4 = 1.

user236182 almost 8 yearsYou must know $\text{ord}_p(x)=8$ in order to conclude $8\mid p1$. And it's not true that $x^k\equiv 1\pmod{p}$ implies either $k\mid p1$ or $p1\mid k$.