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 $p.$

In general, if $p > 2$, $r\ge 2,$ and $n_1,\ldots,n_r$ are integers, then

$$\left(\frac{n_1\cdots n_r}{p}\right)=\left(\frac{n_1}{p}\right)\cdots\left(\frac{n_r}{p}\right).$$

| | | | | created: 2019-05-15 05:33:13 | modified: 2019-05-26 05:28:24 | by: *bookofproofs* | references: [1272]

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[1272] **Landau, Edmund**: “Vorlesungen über Zahlentheorie, Aus der Elementaren Zahlentheorie”, S. Hirzel, Leipzig, 1927