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Fourth and eighth powers modulo $17$

Example

The nonzero fourth powers modulo $17$ are $1, 4, 13, 16$ , and the nonzero eighth powers are $1, 16$ . Each fourth-power target has four fourth roots, and each eighth-power target has eight eighth roots.

Facts & Assumptions

Given: The prime modulus $17$ and nonzero target classes.

[L1]

If $n \geq 1$ admits a primitive root, $\gcd (a, n) = 1$ , $m \geq 1$ , and $d = \gcd (m, φ (n))$ , then $x^{m} \equiv a (m o d n)$ is soluble exactly when $a^{φ (n) / d} \equiv 1 (m o d n)$ (Euler's criterion: if $n$ has a primitive root, $\gcd (a, n) = 1$ , and $m \geq 1$ , then $x^{m} \equiv a (m o d n)$ is solvable if and only if $a^{φ (n) / \gcd (φ (n), m)} \equiv 1 (m o d n)$ ).

[L2]

Every prime admits a primitive root modulo that prime (Every prime modulus admits a primitive root).

[L3]

For every prime $p$ , $φ (p) = p - 1$ ( $φ (1) = 1$ , and $φ (p) = p - 1$ for every prime $p$ ).

[L4]

Under the hypotheses of [L1], a soluble congruence $x^{m} \equiv a (m o d n)$ has exactly $\gcd (φ (n), m)$ solution classes (If $n$ has a primitive root, $\gcd (a, n) = 1$ , $m \geq 1$ , and $x^{m} \equiv a (m o d n)$ is solvable, then it has exactly $\gcd (φ (n), m)$ solution classes modulo $n$ ).

[L5]

If $p ∤ a$ , then $p$ and $a$ are coprime (For a prime $p$ and any integer $a$ , $\gcd (p, a)$ is $p$ when $p ∣ a$ and $1$ otherwise; so $p ∤ a$ makes $p$ and $a$ coprime).

Verification

technique · direct

1.1L1L2L3L5given

By [L2], $17$ admits a primitive root, [L3] gives $φ (17) = 16$ , and [L5] supplies the coprimality of every nonzero target. Thus [L1] says that a target is a fourth power exactly when its fourth power is $1$ , and an eighth power exactly when its square is $1$ .

2.1L1step 1.1algebra

Direct reduction gives the fourth-power values $1^{4} = 4^{4} = 13^{4} = 16^{4} \equiv 1$ , $2^{4} = 8^{4} = 9^{4} = 15^{4} \equiv 16$ , $3^{4} = 5^{4} = 12^{4} = 14^{4} \equiv 13$ , and $6^{4} = 7^{4} = 10^{4} = 11^{4} \equiv 4$ . Squaring the unit classes shows that the targets satisfying $a^{2} \equiv 1$ are exactly $1, 16$ . These lists agree with both directions of the tests in [L1].

3.1L3L4step 2.1algebra∎

Fact [L4] gives $\gcd (16, 4) = 4$ roots for each fourth-power target and $\gcd (16, 8) = 8$ roots for each eighth-power target.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

31 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

A. Gorodnik, Number Theory, Lecture 9, Example 1.3 (standard reference, not scraped)

Fourth and eighth powers modulo 1717