Math 2.0

statement theorem

Quadratic reciprocity for the Jacobi symbol

Statement

Let mm and nn be coprime positive odd integers. Then (mn)(nm)=(−1)m−12⋅n−12,\left(\frac{m}{n}\right) \left(\frac{n}{m}\right) = (-1)^{\frac{m-1}{2} \cdot \frac{n-1}{2}}, where the symbols are Jacobi symbols.

Remarks

The supplements also extend: for a positive odd nn, (−1n)=(−1)(n−1)/2\left(\frac{-1}{n}\right) = (-1)^{(n-1)/2} and (2n)=(−1)(n2−1)/8\left(\frac{2}{n}\right) = (-1)^{(n^2-1)/8}. The proof uses the law of quadratic reciprocity for each pair of prime factors. With this law, one computes (an)\left(\frac{a}{n}\right) by an algorithm like the Euclidean algorithm, with no factorization.

Explanations

L1 human

The Jacobi symbol obeys the same reciprocity law as the Legendre symbol. So you can flip the symbol, reduce the top number modulo the bottom number, and repeat. This gives a fast way to compute Legendre symbols for large primes.

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  1. Divisibility definition
  2. Congruence modulo n definition
  3. Prime number definition
  4. Quadratic residue definition
  5. Fermat's little theorem statement
  6. Prime factorization definition
  7. Legendre symbol definition
  8. Second supplement to quadratic reciprocity statement
  9. Euler's criterion statement
  10. Jacobi symbol definition
  11. First supplement to quadratic reciprocity statement
  12. Law of quadratic reciprocity statement
  13. Quadratic reciprocity for the Jacobi symbol statement

Sources

  1. Web https://en.wikipedia.org/wiki/Jacobi_symbol, Properties