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Difference between revisions of "Legendre symbol"
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==Definition== | ==Definition== | ||
− | If ''p'' is an odd [[prime | + | If ''p'' is an odd [[prime]] number and ''a'' is an [[integer]], then the Legendre symbol |
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− | *[ | + | *[[Wikipedia:Legendre_symbol|Wikipedia]] |
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Revision as of 10:39, 6 February 2019
The Legendre symbol, named after the French mathematician Adrien-Marie Legendre, is used in connection with factorization and quadratic residues.
Definition
If p is an odd prime number and a is an integer, then the Legendre symbol
- [math]\displaystyle{ \left(\frac{a}{p}\right) }[/math]
is:
- 0 if p divides a;
- 1 if a is a square modulo p — that is to say there exists an integer k such that k2 ≡ a (mod p), or in other words a is a quadratic residue modulo p;
- −1 if a is not a square modulo p, or in other words a is not a quadratic residue modulo p.
Properties of the Legendre symbol
There are a number of useful properties of the Legendre symbol which can be used to speed up calculations. They include:
- [math]\displaystyle{ \left(\frac{ab}{p}\right) = \left(\frac{a}{p}\right)\left(\frac{b}{p}\right) }[/math] (it is a completely multiplicative function in its top argument)
- If a ≡ b (mod p), then [math]\displaystyle{ \left(\frac{a}{p}\right) = \left(\frac{b}{p}\right) }[/math]
- [math]\displaystyle{ \left(\frac{1}{p}\right) = 1 }[/math]
- [math]\displaystyle{ \left(\frac{-1}{p}\right) = (-1)^{(p-1)/2} }[/math], i.e. = 1 if p ≡ 1 (mod 4) and = −1 if p ≡ 3 (mod 4)
- [math]\displaystyle{ \left(\frac{2}{p}\right) = (-1)^{(p^2-1)/8} }[/math], i.e. = 1 if p ≡ 1 or 7 (mod 8) and = −1 if p ≡ 3 or 5 (mod 8)
- If q is an odd prime then [math]\displaystyle{ \left(\frac{q}{p}\right) = \left(\frac{p}{q}\right)(-1)^{ ((p-1)/2) ((q-1)/2) } }[/math]
The last property is known as the law of quadratic reciprocity. The properties 4 and 5 are traditionally known as the supplements to quadratic reciprocity.
The Legendre symbol is related to Euler's criterion and Euler proved that
- [math]\displaystyle{ \left(\frac{a}{p}\right) \equiv a^{(p-1)/2}\,\pmod p }[/math]