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Section 4.4 Euler's Theorem

First, let's restate the definition of Euler's totient function (introduced in Exploration 3.2.1).

Definition 4.4.1. Euler's Totient Function.

Euler's totient function (or Euler's phi-function) is the function \(\phi: \mathbb{N} \rightarrow \mathbb{N}\) defined by

\begin{equation*} \phi(n) = \bigl|\{k \in \mathbb{N} : 1 \leq k \leq n, \gcd(k,n) = 1\}\bigr|\text{.} \end{equation*}

Video: Euler's Phi Function

We have \(\phi(1) = 1\) since \(1\) is the only integer in \(\{1\}\) relatively prime with 1.

Also \(\phi(8) = 4\text{,}\) since there are four numbers in the set \(\{1,2,\ldots,8\}\) relatively prime with 8: they are 1, 3, 5, and 7.

The table below lists values of \(\phi(m)\) for small values of \(m\text{.}\) Complete the table for \(m = 11, 12, \ldots, 20\text{.}\)

Table 4.4.4. Values of \(\phi(n)\)
\(n\) integers \(1 \leq k \leq n\) \(\phi(n)\)
such that \(\gcd(k,n)=1\)
1 1 1
2 1 1
3 1, 2 2
4 1, 3 2
5 1, 2, 3, 4 4
6 1, 5 2
7 1, 2, 3, 4, 5, 6 6
8 1, 3, 5, 7 4
9 1, 2, 4, 5, 7, 8 6
10 1, 3, 7, 9 4
11
12
13
14
15
16
17
18
19
20

If \(p\) is prime, then by definition all integers from \(1\) to \(p-1\) are relatively prime with \(p\text{.}\) This implies the following result:

Is the converse of the above statement true? That is, if \(n \gt 2\) is an integer such that \(\phi(n) = n-1\text{,}\) does it necessarily follow that \(n\) is prime?

Justify your answer.

If \(p\) is prime and \(k \in \mathbb{N}\text{,}\) prove that \(\phi(p^k) = p^k - p^{k-1}\text{.}\)

If \(p\) and \(q\) are prime, prove that \(\phi(pq) = (p-1)(q-1)\text{.}\)

Hint

Theorem 3.2.6

Now we can state Euler's Theorem then prove Fermat's Little Theorem, which is a special case.

Video: Two Theorems in Number Theory

If \(p\) is prime with \(p \nmid a\text{,}\) we have \(\gcd(a,p) = 1\text{.}\) So we can use Theorem 4.4.9 on \(a\) and \(m = p\text{.}\) Since \(\phi(p) = p-1\) by Proposition 4.4.5, the result follows.

Although Theorem 4.4.10 is a special case of Theorem 4.4.9, it was Theorem 4.4.10 that was actually proven first—in 1736, also by Euler [1]. Only in 1763 did Euler publish the generalization that is Euler's Theorem [2].

Take \(n = 8\) and \(a = 7\) so that \(\gcd(a,n) = 1\text{.}\) By Theorem 4.4.9 we have

\begin{equation*} 7^{\phi(8)} \equiv 1 \Mod{8} \Rightarrow 7^4 \equiv 1 \Mod{8}\text{.} \end{equation*}

Compute \(3^{2020} \mmod 113\text{.}\)

Solution

Since \(\gcd(3,113) = 1\text{,}\) we can apply Theorem 4.4.9. In fact, 113 is prime, so Theorem 4.4.10 applies here, so we know that

\begin{equation*} 3^{112} \equiv 1 \Mod{13}\text{.} \end{equation*}

Using the Theorem 1.3.2 on 2020 we get \(2019 = 18\cdot 112 + 4\text{.}\) Hence

\begin{equation*} 3^{2020} \equiv (3^{112})^{18} \cdot 3^4 \Mod{113} \equiv 81 \Mod{113}\text{.} \end{equation*}

So \(3^{2020} \mmod 113 = 81\text{.}\)

Compute \(6^{6777} \mmod 667\text{.}\)

Hint

\(667 = 23 \times 29.\)

The idea behind the proof of Theorem 4.4.9 is that for \(a,n \in \mathbb{N}\) relatively prime, multiplying each element of the set

\begin{equation*} S = \{y \in \{1,2,\ldots,n\} : \gcd(y,n) = 1\} \end{equation*}

by \(a\) induces a permutation of the set modulo \(n\text{.}\) We give first an example with numbers.

Let \(a = 4\) and \(n = 9\text{.}\) Then \(\phi(9) = 6\text{,}\) since the integers \(S = \{1,2,4,5,7,8\}\) are relatively prime with \(9\text{.}\)

Multiplying each number in \(S\) by 4 and reducing modulo 9, we have

\(S = \) \(\{\)1 2 4 5 7 8\(\}\)
\(\downarrow\) \(\downarrow\) \(\downarrow\) \(\downarrow\) \(\downarrow\) \(\downarrow\) (multiply by \(a = 4\))
4 8 16 20 28 32
\(\downarrow\) \(\downarrow\) \(\downarrow\) \(\downarrow\) \(\downarrow\) \(\downarrow\) (reduce modulo 9)
\(\{\)4 8 7 2 1 5\(\}\) \(\rightarrow\) same as \(S\text{!}\)

What this means is that \(S\) and \(\{4y : y \in S\}\) contain the same numbers modulo 9; so when we take the product of all elements in these sets, the results must be congruent modulo 9 as well:

\begin{equation*} (4\cdot 1)(4\cdot 2)(4\cdot 4)(4 \cdot 5)(4\cdot 7)(4\cdot 8) \equiv (1)(2)(4)(5)(7)(8)\Mod{9}\text{.} \end{equation*}

Since each element of \(S\) is relatively prime with 9, we can apply Proposition 4.2.15 to cancel each term, and we are left with

\begin{equation*} (4)(4)(4)(4)(4)(4) \equiv 1 \Mod{9} \Rightarrow 4^{\phi(9)} \equiv 1 \Mod{9} \end{equation*}

since \(\phi(9) = 6 = |S|\text{.}\)

Replicate the idea in Example 4.4.14 to show that \(7^{\phi(15)} \equiv 1 \Mod{15}\text{.}\)

Finally, we present the proof of Theorem 4.4.9.

Proof of Euler's Theorem

Let \(a,n \in \mathbb{N}\) such that \(\gcd(a,n) = 1\text{.}\)

First define the set

\begin{equation*} S = \{y : 1 \leq y \leq n, \gcd(y,n) = 1\} \end{equation*}

to be the set of natural numbers smaller than \(n\) and relatively prime with \(n\text{.}\) We know that there are exactly \(\phi(n)\) elements in the set \(S\text{,}\) so we can label them as \(S = \{b_1,b_2,\ldots,b_{\phi(n)}\}\text{,}\) where \(\gcd(b_i,n) = 1\) for \(i = 1,2,\ldots,\phi(n)\text{.}\)

Since we have \(\gcd(b_i,n) = 1\) and \(\gcd(a,n) = 1\text{,}\) we must also have

\begin{equation*} \gcd(ab_i, n) = 1 \text{ for any } i = 1,2,\ldots,\phi(n)\text{.} \end{equation*}

We invoke Theorem 1.3.2 now to write \(ab_i\) as

\begin{equation*} ab_i = qn + r \text{ for some } 0 \leq r \lt n, \end{equation*}

which implies that \(ab_i \equiv r \Mod{n}\text{.}\)

Combining with the fact that \(\gcd(ab_i,n) = 1\text{,}\) this implies that \(\gcd(r,n) = 1\text{,}\) so \(r\) is a natural number smaller than \(n\) and relatively prime with \(n\text{.}\)

In other words, \(r\) is in the set \(S\text{,}\) and we can write \(r = b_j\) for some \(1 \leq j \leq \phi(n)\text{.}\) This means for each \(1 \leq i \leq \phi(n)\text{,}\) we have \(ab_i \equiv b_j \Mod{n}\) for some \(1 \leq j \leq \phi(n)\text{.}\)

Furthermore, none of the \(b_j\)'s are repeated, because no two \(ab_i\) terms are equivalent modulo \(n\text{.}\) (Otherwise, \(ab_{i_1} \equiv ab_{i_2} \Mod{n} \Rightarrow b_{i_1} \equiv b_{i_2} \Mod{n}\) by Proposition 4.2.15, which is a contradiction as the \(b_i\)'s are all distinct and smaller than \(n\text{.}\))

Hence, each integer \(ab_i\) is congruent modulo \(n\) to distinct elements in \(S\text{:}\)

\begin{align*} ab_1 \amp \equiv b_{j_1} \Mod{n}\\ ab_2 \amp \equiv b_{j_2} \Mod{n}\\ \vdots \amp\\ ab_{\phi(n)} \amp \equiv b_{j_{\phi(n)}} \Mod{n} \end{align*}

Multiplying all the congruences together gives another congruence, where the right-hand-side is just

\begin{equation*} \displaystyle\prod_{k=1}^{\phi(n)} b_{j_k} = \displaystyle\prod_{i=1}^{\phi(n)} b_i \end{equation*}

since each element of \(S\) appears exactly once. Hence,

\begin{equation*} \prod_{i=1}^{\phi(n)} ab_i \equiv \prod_{i=1}^{\phi(n)} b_i \Mod{n} \Rightarrow a^{\phi(n)} \equiv 1 \Mod{n} \end{equation*}

by an application of Proposition 4.2.15.

Video: Primality Testing

Exploration 4.4.1. Primality Testing.

Theorem 4.4.10 asserts that if \(p\) is prime and \(\gcd(a,p)=1\text{,}\) then \(a^{p-1} \equiv 1 \Mod{p}\text{.}\)

(a)

Note that the converse of this statement is not true in general: Even if \(\gcd(a,n) = 1\) and \(a^{n-1} \equiv 1 \Mod{n}\text{,}\) we would not be able to conclude that \(n\) is prime.

Can you give examples of pairs \(a, n\) such that \(\gcd(a,n) = 1\) and \(a^{n-1} \equiv 1 \Mod{n}\) are both true, but \(n\) is not prime.

(b)

Complete the contrapositive of Theorem 4.4.10:

If \(a,n \in \mathbb{N}\text{,}\) \(\gcd(a,n) = 1\text{,}\) and \(a^{n-1} \not\equiv 1 \Mod{n}\text{,}\) then .

(c)

Show that 91 is not prime by using part (b) with \(a = 2\text{.}\)