Difference between revisions of "Quadratic Reciprocity Theorem"

m (Statement)
m (Statement)
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==Statement==
 
==Statement==
It states that <math>\left(\frac{p}{q}\right)= \left(\frac{q}{p}\right)</math>* for primes <math>p</math> and <math>q</math> greater than <math>2</math> where both are not of the form <math>4n+3</math> for some integer <math>n</math>.<br>
+
It states that <math>\left(\frac{p}{q}\right)= \left(\frac{q}{p}\right)</math> for primes <math>p</math> and <math>q</math> greater than <math>2</math> where both are not of the form <math>4n+3</math> for some integer <math>n</math>.<br>
 
If both <math>p</math> and <math>q</math> are of the form <math>4n+3</math>, then <math>\left(\frac{p}{q}\right)= -\left(\frac{q}{p}\right).</math>
 
If both <math>p</math> and <math>q</math> are of the form <math>4n+3</math>, then <math>\left(\frac{p}{q}\right)= -\left(\frac{q}{p}\right).</math>
  
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<math>\left(\frac{p}{q}\right)\left(\frac{q}{p}\right)=(-1)^{\frac{p-1}{2}\frac{q-1}{2}}.</math>
 
<math>\left(\frac{p}{q}\right)\left(\frac{q}{p}\right)=(-1)^{\frac{p-1}{2}\frac{q-1}{2}}.</math>
  
*Note that <math>\left(\frac{p}{q}\right)</math> is [b]not[/b] a fraction. It is the [b]Legendre notation[/b] of quadratic residuary.
+
Note that <math>\left(\frac{p}{q}\right)</math> is not a fraction. It is the Legendre notation of quadratic residuary.
  
 
==See Also==
 
==See Also==
 
[[Category:Number theory]]
 
[[Category:Number theory]]
 
[[Category:Theorems]]
 
[[Category:Theorems]]

Revision as of 23:19, 5 April 2021

Quadratic reciprocity is a classic result of number theory.
It is one of the most important theorems in the study of quadratic residues.

Statement

It states that $\left(\frac{p}{q}\right)= \left(\frac{q}{p}\right)$ for primes $p$ and $q$ greater than $2$ where both are not of the form $4n+3$ for some integer $n$.
If both $p$ and $q$ are of the form $4n+3$, then $\left(\frac{p}{q}\right)= -\left(\frac{q}{p}\right).$

Another way to state this is:
$\left(\frac{p}{q}\right)\left(\frac{q}{p}\right)=(-1)^{\frac{p-1}{2}\frac{q-1}{2}}.$

Note that $\left(\frac{p}{q}\right)$ is not a fraction. It is the Legendre notation of quadratic residuary.

See Also