Art of Problem Solving

Quadratic Reciprocity Theorem: Difference between revisions

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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 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]]
[[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