CorollaryStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-07-31

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

Statement

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

$\varphi(p)=p-1.$

Facts & Assumptions

Given: A prime integer $p$ .

[F1]

$\varphi(n)$ is the cardinality of the unit group of $\mathbb Z/n$ for positive $n$ , and a class is a unit exactly when its representative is coprime to $n$ (The unit group $(\mathbb{Z}/n)^\times$ and Euler's totient $\varphi(n)=\lvert(\mathbb{Z}/n)^\times\rvert$ for $n\ge1$ , For $n\ge1$ , $[a]_n$ is a unit if and only if $\gcd(a,n)=1$ ).

[L1]

For a prime $p$ , if $p\nmid a$ then $\gcd(p,a)=1$ (For a prime $p$ and any integer $a$ , $\gcd(p,a)$ is $p$ when $p \mid a$ and $1$ otherwise; so $p \nmid a$ makes $p$ and $a$ coprime).

[L2]

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

[L3]

A subset of a finite set is finite (A subset of a finite set is finite, with $\lvert B\rvert \le \lvert A\rvert$ , 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 $\varphi(1)=1$ .

F1L2

1.2

Let $0\le r<p$ . If $r=0$ , then $p\mid r$ , so $[r]_p$ is not a unit by the unit criterion. If $r>0$ , then $p\nmid r$ : otherwise $r=pk$ with $p>0$ and $r>0$ , forcing $k\ge1$ and $r\ge p$ , 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 $1,2,\ldots,p-1$ .

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, $\varphi(p)=p-1$ ; together with step 1.1 this proves both clauses.

step 1.1step 3.1F1∎

Depends on

Used by

Fermat's little theorem: for prime p, p∤ a implies aᵖ⁻¹≡1pmod p, and always aᵖ≡ apmod p Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 83 results over 24 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

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

φ(1)=1\varphi(1)=1φ(1)=1, and φ(p)=p−1\varphi(p)=p-1φ(p)=p−1 for every prime ppp