CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-07-31

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

Statement

Euler's totient satisfies $φ (1) = 1$ . If $p$ is prime (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ), then

$φ (p) = p - 1.$

Facts & Assumptions

Given: A prime integer $p$ .

[F1]

$φ (n)$ is the cardinality of the unit group of $Z / n$ for positive $n$ , and a class is a unit exactly when its representative is coprime to $n$ (The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ , For $n \geq 1$ , $[a]_{n}$ is a unit if and only if $\gcd (a, n) = 1$ ).

[L1]

For a prime $p$ , if $p ∤ a$ then $\gcd (p, a) = 1$ (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).

[L2]

The standard representatives modulo a positive $n$ are the unique integers $r$ with $0 \leq r < n$ , and there are $n$ of them (For $n \geq 1$ , every class in $Z / n$ has one representative $r$ with $0 \leq r < n$ , so $∣ Z / n ∣ = n$ ; while $Z / 0$ is in bijection with $Z$ , The cardinality $∣ A ∣$ of a finite set).

[L3]

A subset of a finite set is finite (A subset of a finite set is finite, with $∣ B ∣ \leq ∣ A ∣$ , and equality holds if and only if $B = A$ ).

[F2]

A prime integer satisfies $p > 1$ (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ).

Proof

technique · direct

1.1

Modulo $1$ there is one class, and it is the multiplicative identity, hence a unit. Therefore $φ (1) = 1$ .

F1L2

1.2

Let $0 \leq r < p$ . If $r = 0$ , then $p ∣ r$ , so $[r]_{p}$ is not a unit by the unit criterion. If $r > 0$ , then $p ∤ r$ : otherwise $r = p k$ with $p > 0$ and $r > 0$ , forcing $k \geq 1$ and $r \geq p$ , contrary to $r < p$ .

F1F2L2algebra

2.1

For $0 < r < p$ , [L1] gives $\gcd (p, r) = 1$ , hence $\gcd (r, p) = 1$ , so $[r]_{p}$ is a unit by the unit criterion. Thus the units are exactly the classes with representatives $1, 2, \dots, p - 1$ .

step 1.2F1L1

3.1

Translation by $1$ is a bijection from the natural $p - 1$ onto the representatives $r$ with $0 < r < p$ . Hence that finite set, and therefore the unit group, has cardinality $p - 1$ .

step 2.1L2L3algebra

4.1

By [F1] and step 3.1, $φ (p) = p - 1$ ; together with step 1.1 this proves both clauses.

step 1.1step 3.1F1∎

Depends on

Used by

Dependency tree · two levels

45 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

Mathematics LibreTexts, Euler's phi Function (standard reference, not scraped)
UW Oshkosh, Number Theory Notes 3 (standard reference, not scraped)

φ(1)=1φ(1)=1, and φ(p)=p−1φ(p)=p−1 for every prime pp