2
  • It seems that for every prime p = (4k + 1), p-1 is a quadratic residue modulo p.
  • It seems that for every prime p = (4k - 1), p-1 is not a quadratic residue modulo p.

I can't figure how to proove this, does anyone has an idea ?

Best regards

BenLaz
  • 817
  • 1
    Ever heard of Euler's criterion? – Wojowu Dec 20 '16 at 08:32
  • http://math.stackexchange.com/questions/124688/number-of-quadratic-residues-mod-p – lab bhattacharjee Dec 20 '16 at 08:36
  • Thank you for your answer, but i can't see how it leads to a proof. – BenLaz Dec 20 '16 at 09:05
  • 1
    This has been explained many time on the site. You will probably have more luck searching for when $-1$ is a quadratic residue (as opposed to the congruent $p-1$). Some common arguments: A) If $a^2\equiv-1$ then $a$ is of order $4$ in $\Bbb{Z}_p^$, so by Lagrange $4\mid p-1$. B) If $p\equiv1\pmod4$, then $a=((p-1)/2)!$ has the property $a^2\equiv-1\pmod p$. C) The group $\Bbb{Z}_p^$ is cycli of order $p-1$, so if $4\mid p-1$ there is an element of order four ... Pick and choose :-) – Jyrki Lahtonen Dec 20 '16 at 09:28
  • Good observation. I touch on this in my question about primitive roots http://math.stackexchange.com/questions/1186548/count-of-lower-and-upper-primitive-roots-of-prime-p-equiv-3-bmod-4 – Joffan Dec 20 '16 at 09:29

2 Answers2

1

Your question is: For $ p \equiv 3 \mod 4$, $p-1$ is not a quadratic residue modulo $p$.

Recall Euler's Criterion:

$a$ is a quadratic residue if and only if $a^{\frac{p-1}{2}} \equiv 1 \mod p$. (I can supply a proof of this if necessary).

Now for the case of $p-1$ we see that $p-1 \equiv -1 \mod p$, so our original question is equivalent to asking whether $-1^{\frac{p-1}{2}} \equiv 1 \mod p$. Notice that if $p \equiv 3 \mod 4$.

Then $p - 1 \equiv 2 \mod 4$, and $\frac{p-1}{2} \equiv 1 \mod 4$. In particular $\frac{p-1}{2}$ is odd.

Thus $(-1)^{\frac{p-1}{2}} = -1$. Thus ${(p-1)}^{\frac{p-1}{2}} \equiv (-1)^{\frac{p-1}{2}} \equiv -1 \mod p$. Thus, by Euler's Criterion, $p-1$ is not a quadratic residue modulo $p$ if $p \equiv 3 \mod 4$.

A similar argument shows that if $p \equiv 1 \mod 4$, $p-1$ is a quadratic residue modulo $p$.

Sean Haight
  • 1,368
  • 11
  • 26
0

Every odd prime $p$ has a primitive root $g$ (actually many) such that the smallest $k>o$ for which $g^k \equiv 1 \bmod p$ is $k=p-1$.

In that case, $g^{(p\mathord-1)/2} \equiv -1 \bmod p$. If $(p\mathord-1)/2$ is even then there is a value $a\equiv g^{(p\mathord-1)/4}$ such that $a^2\equiv -1 \bmod p$ and $-1$ is a quadratic residue $\bmod p$. If $(p\mathord-1)/2$ is odd, there is no such number and $-1$ is not a quadratic residue $\bmod p$. The parity of $(p\mathord-1)/2$ translates into a statement about whether $1$ or $3 \equiv p \bmod 4$.

Joffan
  • 40,356