• Jun 19th 2011, 02:19 PM
swashbucklord
Hi all, messed up using Latex at first. Okay, got two problems on quadratic reciprocity. I've worked up something of a proof for each, but I feel very uncertain. Thought maybe someone could give me some pointers.

Okay, simple one first:
Suppose p is a prime such that $p\equiv 3 (mod 4)$, and let a be a QR modulo p.
Show that $x=a^\frac{p+1}{4}$ is a solution to
$x^2\equiv a(mod\, p)$

My proof:
a - QR mod p $\Rightarrow$ the Legendre symbol
$(\frac{a}{p})= 1 \equiv a^{\frac{p-1}{2}}$,
so clearly $1\equiv a^{\frac{p-1}{2}} (mod\, p)$.
Multiplying both sides by a yields
$a\equiv a^{\frac{p-1}{2}}*a\equiv a^{\frac{p+1}{2}}$.
But if $x=a^{\frac{p+1}{4}}$,
we have that $x^{2}=a^{\frac{p+1}{2}}$.
We know that $a^{\frac{p+1}{2}}$ is congruent to a modulo p, so
$x=a^{\frac{p+1}{4}}$ is a solution to the congruence.

My trouble is that this seems too easy, and not once did I apply $p\equiv 3(mod\, 4)$.

The 'solution' $x= a^{\frac{p+1}{4}}$ must of course be an integer and that happens only if $p \equiv 3\ (\text{mod}\ 4)$...
$\chi$ $\sigma$