Quadratic Reciprocity Proof (working now)

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 $\displaystyle p\equiv 3 (mod 4)$, and let a be a QR modulo p.

Show that $\displaystyle x=a^\frac{p+1}{4}$ is a solution to

$\displaystyle x^2\equiv a(mod\, p)$

My proof:

a - QR mod p $\displaystyle \Rightarrow $ the Legendre symbol

$\displaystyle (\frac{a}{p})= 1 \equiv a^{\frac{p-1}{2}}$,

so clearly $\displaystyle 1\equiv a^{\frac{p-1}{2}} (mod\, p)$.

Multiplying both sides by a yields

$\displaystyle a\equiv a^{\frac{p-1}{2}}*a\equiv a^{\frac{p+1}{2}}$.

But if $\displaystyle x=a^{\frac{p+1}{4}}$,

we have that $\displaystyle x^{2}=a^{\frac{p+1}{2}}$.

We know that $\displaystyle a^{\frac{p+1}{2}}$ is congruent to a modulo p, so

$\displaystyle 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 $\displaystyle p\equiv 3(mod\, 4)$.

Any pointers? Thanks in advance.

Re: Quadratic Reciprocity Proof (working now)

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$