Proposition: Multiplicativity of the Legendre Symbol
Let $p > 2$ be a prime number.
In other words:
- The congruence $x^2(p)\equiv nm(p)$ is solvable if and only if the congruences $x^2(p)\equiv n(p)$ and $x^2(p)\equiv m(p)$ are solvable.
Yet in other words:
- If $n$ and $m$ are both quadratic residues (respectively both quadratic nonresidues) modulo $p$, then their product $nm$ is a quadratic residue modulo $p.$
- If one of the numbers $n$ and $m$ is a quadratic residue modulo $p$ and the other a quadratic nonresidue modulo $p$, then their product $nm$ is a quadratic nonresidue modulo $p.$
with $p\not\mid n$ and $p\not\mid m.$
Then the Legendre symbols of $m,n$ modulo $p$
| | | | | created: 2019-05-15 05:33:13 | modified: 2019-05-15 05:34:10 | by: bookofproofs | references: 
1.Proof: (related to "Multiplicativity of the Legendre Symbol")
This work is a derivative of:
Bibliography (further reading)
 Landau, Edmund: “Vorlesungen über Zahlentheorie, Aus der Elementaren Zahlentheorie”, S. Hirzel, Leipzig, 1927