Difference between revisions of "Quadratic Reciprocity Theorem"

 
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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>
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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>
  
 
Another way to state this is:<br>
 
Another way to state this is:<br>
<math>\left(\frac{p}{q}\right)\left(\frac{q}{p}\right)=(-1)^{\frac{p-1}{2}\frac{q-1}{2}}</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>
 
 
==Links==
 
Quadratic Residues[http://www.artofproblemsolving.com/Wiki/index.php/Quadratic_residues]
 
  
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*Note that <math>\left(\frac{p}{q}\right)</math> is not a fraction. It is the Legendre notation of quadratic residuary.
  
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==See Also==
 
[[Category:Number theory]]
 
[[Category:Number theory]]
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[[Category:Theorems]]
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[[Quadratic Residues |Quadratic Residues]]

Latest revision as of 22:21, 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

Quadratic Residues