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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

1.1

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 ${1, 4}$ .

L2L3givenalgebra

2.1

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

L1L2step 1.1

3.1

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

step 2.1algebra∎

Depends on

Used by

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Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 42 results over 16 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click 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