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.
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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.
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