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## Proposition: Multiplicativity of the Legendre Symbol

Let $p > 2$ be a prime number.

The Legendre symbols are completely multiplicative, i.e. for any two integers $n,m\in\mathbb Z$ we have $$\left(\frac{nm}{p}\right)=\left(\frac{n}{p}\right)\cdot\left(\frac{m}{p}\right).$$

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: [1272]

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