Thread: 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.

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

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).

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.

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

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

Here are the four cases:

and
and
and
and

I'll verify one case for you:
and

We know .
Therefore

We're also given that and .
So .
Thus .

(Notice that is proves the case and too since and are arbitrary.