Difference between revisions of "Quadratic reciprocity"
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* <math>\left(\frac{-1}{p}\right)=(-1)^{(p-1)/4}</math>. | * <math>\left(\frac{-1}{p}\right)=(-1)^{(p-1)/4}</math>. | ||
− | * <math>\left(\frac{2}{p}\right)=(-1)^{(p^2-1)/8 | + | * <math>\left(\frac{2}{p}\right)=(-1)^{(p^2-1)/8}</math>. |
* <math>\left(\frac{p}{q}\right)\left(\frac{q}{p}\right)=(-1)^{(p-1)/4\ (q-1)/4}</math>. | * <math>\left(\frac{p}{q}\right)\left(\frac{q}{p}\right)=(-1)^{(p-1)/4\ (q-1)/4}</math>. | ||
Revision as of 17:14, 12 July 2006
Let be a prime, and let
be any integer not divisible by
. Then we can define the Legendre symbol $\left(\frac{a}{p}\right)=
is a quadratic residue modulo
if there exists an integer
so that
. We can then define
if
is divisible by
.
Quadratic Reciprocity Theorem
There are three parts. Let and
be distinct odd primes. The the following hold:
.
.
.
This theorem can help us evaluate Legendre symbols, since the following laws also apply:
- If
, then
.
.
There also exist quadratic reciprocity laws in other rings of integers. (I'll put that here later if I remember.)