1. ## Quadratic congruence - proof

Show that the quadratic reciprocity theorem could also be written (p/q)(q/p) = (-1)^[(p-1)(q-1)/4] for odd primes p and q.

2. That's usually how it's written - which form do you have on hand?

3. My text uses the following:

If p and q are odd primes and p==q==3(mod 4), then (p/q) = -(q/p). Otherwise, (p/q) = (q/p).

4. Originally Posted by tarheelborn
My text uses the following:

If p and q are odd primes and p==q==3(mod 4), then (p/q) = -(q/p). Otherwise, (p/q) = (q/p).
Well, there's four cases to verify, so a quick case by case proof should do the trick.

5. p==q==3(mod 4), then (p/q) = -(q/p).

Meaning p==3(mod 4), q==3(mod 4), p==q(mod 4) and?

6. Here are the four cases:

$\displaystyle p\equiv1\mod{4}$ and $\displaystyle q\equiv1\mod{4}$
$\displaystyle p\equiv1\mod{4}$ and $\displaystyle q\equiv3\mod{4}$
$\displaystyle p\equiv3\mod{4}$ and $\displaystyle q\equiv1\mod{4}$
$\displaystyle p\equiv3\mod{4}$ and $\displaystyle q\equiv3\mod{4}$

I'll verify one case for you:
$\displaystyle p\equiv1\mod{4}$ and $\displaystyle q\equiv3\mod{4}$

We know $\displaystyle \left(\frac{p}{q}\right) = \left(\frac{q}{p}\right)$.
Therefore $\displaystyle \left(\frac{p}{q}\right)\left(\frac{q}{p}\right) = \left(\frac{p}{q}\right)^2 = (\pm1)^2 = 1$

We're also given that $\displaystyle p=4k+1$ and $\displaystyle q=4n+3$.
So $\displaystyle \tfrac14(p-1)(q-1) = 2k(2n+1) = 2x$.
Thus $\displaystyle (-1)^{\frac{(p-1)(q-1)}{4}} = (-1)^{2x} = 1$.

(Notice that is proves the case $\displaystyle p\equiv3\mod{4}$ and $\displaystyle q\equiv1\mod{4}$ too since $\displaystyle p$ and $\displaystyle q$ are arbitrary.