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Two quadratic nonresidues modulo $15$ can have a nonresidue product

Statement refuted

The product of two quadratic nonresidues modulo an integer need not be a quadratic residue. Modulo $15$ , the classes of $2$ and $7$ are nonresidues, and their product $14$ is also a nonresidue.

Facts & Assumptions

Given: The composite modulus $15$ .

[L1]

A unit integer is a quadratic residue modulo $n$ exactly when it has a square root modulo $n$ , and otherwise it is a quadratic nonresidue (Quadratic residues and nonresidues modulo an integer).

[L2]

The quadratic-residue classes are exactly the image of squaring on $(Z / n)^{\times}$ (Quadratic residuosity is representative-independent and the residues are the image of squaring).

[L3]

A class $[a]_{n}$ is a unit exactly when $\gcd (a, n) = 1$ (For $n \geq 1$ , $[a]_{n}$ is a unit if and only if $\gcd (a, n) = 1$ ).

Counterexample

technique · direct

1.1L2L3givenalgebra

By [L3], the units modulo $15$ are $1, 2, 4, 7, 8, 11, 13, 14$ . Squaring them gives respectively $1, 4, 1, 4, 4, 1, 4, 1$ , so [L2] identifies the square image as exactly ${1, 4}$ .

2.1L1L2step 1.1

The unit classes $2, 7, 14$ lie outside that image, so [L1] and [L2] make all three quadratic nonresidues modulo $15$ .

3.1step 2.1algebra∎

Yet $2 \cdot 7 = 14 (m o d 15)$ , so the product of the two nonresidues $2$ and $7$ is the nonresidue $14$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

12 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

H. Hackman, Elementary Number Theory, Chapter D, Section D.I (standard reference, not scraped)

Two quadratic nonresidues modulo 1515 can have a nonresidue product