-1 as a Quadratic Residue mod $p$ $\Rightarrow$ $p \equiv_4 1$

Question

Suppose $p$ is odd prime.

If $x^2 \equiv_p -1$, show $(x^2)^{\frac{p-1}{2}} \equiv_p 1$, and conclude that $p \equiv_4 1$ ( I cannot get to this part for some stupid reason)

Here is what I have,

We wish to show

$(x^2)^{\frac{p-1}{2}} \equiv_p 1$

if we square both sides we have

$(x^2)^{p-1} \equiv_p 1$ **

but $x^2 \equiv_p -1$ hence ** becomes

$(-1)^{p-1} \equiv_p 1$ and $p - 1$ is even, we get

$1 \equiv_p 1$ which clearly holds.

Now how do I go from assuming

$(x^2)^{\frac{p-1}{2}} \equiv_p 1$

and concluding that $p \equiv_4 1$?

I tried saying $p \vert (x^{\frac{p-1}{2}} - 1)(x^{\frac{p-1}{2}} + 1)$

and since $p$ is prime is has to divide one of the two?

$x^2\equiv-1$ ; $(-1)^{even}=1; (-1)^{odd}=-1$ so $(x^2)^{\frac{p-1}2}\equiv 1 $ if $\frac{p-1}2$ is even and $-1$ if $\frac{p-1}2$ is odd — J. W. Tanner, May 15 '19 at 00:54

score 1 · Accepted Answer · answered May 15 '19 at 01:15

Let $p$ be an odd prime and assume $x^2\equiv-1\mod p$.

Then $x^{p-1}=(x^2 )^\frac{p-1}2\equiv(-1)^\frac{p-1}2=1 $ if $\frac{p-1}2 $ is even (i.e., $p\equiv1\mod 4$)

and $-1$ if $\frac{p-1}2$ is odd (i.e., $p\equiv3\mod4$).

But $x$ is not divisible by $p$ (otherwise we'd have $x^2\equiv0\mod p$), so, by Fermat's little theorem,

$x^{p-1}\equiv1\mod p $. Therefore $p\equiv1\mod 4$.

1 Answers1