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Difference between revisions of "Euler's Totient Theorem"

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== Theorem ==
 
== Theorem ==
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Let <math>\phi(n)</math> be [[Euler's totient function]]. If <math>{a}</math> is an integer and <math>m</math> is a positive integer [[relatively prime]] to <math>a</math>, then <math>{a}^{\phi (m)}\equiv 1 \pmod {m}</math>.
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Let <math>\phi(n)</math> be [[Euler's totient function]]. If <math>n</math> is a positive integer, <math>\phi{(n)}</math> is the number of integers in the range <math>\{1,2,3\cdots{,n}\}</math> which are relatively prime to <math>n</math>. If <math>{a}</math> is an integer and <math>m</math> is a positive integer [[relatively prime]] to <math>a</math>, then <math>{a}^{\phi (m)}\equiv 1 \pmod {m}</math>.
  
 
== Credit ==
 
== Credit ==
  
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This theorem is credited to [[Leonhard Euler]].  It is a generalization of [[Fermat's Little Theorem]], which specifies that <math>{m}</math> is prime. For this reason it is also known as Euler's generalization or the Fermat-Euler theorem.
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This theorem is credited to [[Leonhard Euler]].  It is a generalization of [[Fermat's Little Theorem]], which specifies it when <math>{m}</math> is prime. For this reason it is also known as Euler's generalization or the Fermat-Euler theorem.
  
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==Proof==
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==Direct Proof==
  
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Consider the set of numbers <math>A = </math>{<math>n_1, n_2, ... n_{\phi(m)} </math>} (mod m) such that the elements of the [[set]] are the numbers relatively [[prime]] to each other.  
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Consider the set of numbers <math>A = \{ n_1, n_2, ... n_{\phi(m)} \} \pmod{m}</math> such that the elements of the [[set]] are the numbers relatively [[prime]] to <math>m</math>.  
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It will now be proved that this set is the same as the set  <math>B = </math>{<math>an_1, an_2, ... an_{\phi(m)} </math>} (mod m) where <math> (a, m) = 1</math>. All elements of <math>B</math> are relatively prime to <math>m</math> so if all elements of <math>B</math> are distinct, then <math>B</math> has the same elements as <math>A</math>. This means that <math> n_1 n_2 ... n_{\phi(m)} \equiv an_1 \cdot an_2 ... an_{\phi(m)}</math>(mod m) → <math>a^{\phi (m)} \cdot (n_1 n_2 ... n_{\phi(m)}) \equiv n_1 n_2 ... n_{\phi(m)}</math> (mod m) → <math>a^{\phi (m)} \equiv 1</math> (mod m) as desired.
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It will now be proved that this set is the same as the set  <math>B = \{ an_1, an_2, ... an_{\phi(m)} \} \pmod{m}</math> where <math>\gcd(a, m) = 1</math>. All elements of <math>B</math> are relatively prime to <math>m</math> so if all elements of <math>B</math> are distinct, then <math>B</math> has the same elements as <math>A.</math> In other words, each element of <math>B</math> is congruent to one of <math>A</math>. This means that <math> n_1 n_2 ... n_{\phi(m)} \equiv an_1 \cdot an_2 ... an_{\phi(m)} \pmod{m} </math> <math> \implies </math> <math> a^{\phi (m)} \cdot (n_1 n_2 ... n_{\phi(m)}) \equiv n_1 n_2 ... n_{\phi(m)} \pmod{m} </math> <math> \implies </math> <math> a^{\phi (m)} \equiv 1 \pmod{m}</math> as desired. Note that dividing by <math> n_1 n_2 ... n_{\phi(m)}</math> is allowed since it is relatively prime to <math>m</math>. <math>\square</math>
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==Group Theoretic Proof==
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Define the set <math>(\mathbb{Z}/n\mathbb{Z})^{\times}=\{a\in\mathbb{N}\,|\,1\le a <n \wedge \gcd(a,n)=1\}</math>. Trivially we have that <math>|(\mathbb{Z}/n\mathbb{Z})^{\times}|=\varphi(n)</math>. From there, there exists a sufficient <math>k</math> such that <math>a^{k}\equiv 1\pmod{n}</math>, and by [[Lagrange's Theorem]] <math>bk|\varphi(n)</math> which means
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<math>a^{\varphi(n)}\equiv a^{bk} \equiv (a^{k})^b\equiv 1^b\equiv 1\pmod{n} </math>, completing the proof. <math>\square</math>
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==Problems==
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===Introductory===
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*(BorealBear) Find the last two digits of <math> 7^{81}-3^{81} </math>. ([[Euler's Totient Theorem Problem 1 Solution|Solution]])
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*(BorealBear) Find the last two digits of <math> 3^{3^{3^{3}}} </math>. ([[Euler's Totient Theorem Problem 2 Solution|Solution]])
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*([[2018 AMC 10B Problems/Problem 16|2018 AMC 10B]]) Let <math>a_1,a_2,\dots,a_{2018}</math> be a strictly increasing sequence of positive integers such that <cmath>a_1+a_2+\cdots+a_{2018}=2018^{2018}.</cmath>
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What is the remainder when <math>a_1^3+a_2^3+\cdots+a_{2018}^3</math> is divided by <math>6</math>?
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 +
<cmath>\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4</cmath>
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*([[1983 AIME Problems/Problem 6|1983 AIME]]) Let <math>a_n=6^{n}+8^{n}</math>. Determine the remainder upon dividing <math>a_ {83}</math> by <math>49</math>.
  
 
== See also ==
 
== See also ==
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* [[Number theory]]
 
* [[Number theory]]
 
* [[Modular arithmetic]]
 
* [[Modular arithmetic]]
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* [[Lagrange's Theorem]]
 
* [[Euler's totient function]]
 
* [[Euler's totient function]]
 
* [[Carmichael function]]
 
* [[Carmichael function]]

Latest revision as of 16:45, 21 March 2023

Euler's Totient Theorem is a theorem closely related to his totient function.

Theorem

Let $\phi(n)$ be Euler's totient function. If $n$ is a positive integer, $\phi{(n)}$ is the number of integers in the range $\{1,2,3\cdots{,n}\}$ which are relatively prime to $n$. If ${a}$ is an integer and $m$ is a positive integer relatively prime to $a$, then ${a}^{\phi (m)}\equiv 1 \pmod {m}$.

Credit

This theorem is credited to Leonhard Euler. It is a generalization of Fermat's Little Theorem, which specifies it when ${m}$ is prime. For this reason it is also known as Euler's generalization or the Fermat-Euler theorem.

Direct Proof

Consider the set of numbers $A = \{ n_1, n_2, ... n_{\phi(m)} \} \pmod{m}$ such that the elements of the set are the numbers relatively prime to $m$. It will now be proved that this set is the same as the set $B = \{ an_1, an_2, ... an_{\phi(m)} \} \pmod{m}$ where $\gcd(a, m) = 1$. All elements of $B$ are relatively prime to $m$ so if all elements of $B$ are distinct, then $B$ has the same elements as $A.$ In other words, each element of $B$ is congruent to one of $A$. This means that $n_1 n_2 ... n_{\phi(m)} \equiv an_1 \cdot an_2 ... an_{\phi(m)} \pmod{m}$ $\implies$ $a^{\phi (m)} \cdot (n_1 n_2 ... n_{\phi(m)}) \equiv n_1 n_2 ... n_{\phi(m)} \pmod{m}$ $\implies$ $a^{\phi (m)} \equiv 1 \pmod{m}$ as desired. Note that dividing by $n_1 n_2 ... n_{\phi(m)}$ is allowed since it is relatively prime to $m$. $\square$

Group Theoretic Proof

Define the set $(\mathbb{Z}/n\mathbb{Z})^{\times}=\{a\in\mathbb{N}\,|\,1\le a <n \wedge \gcd(a,n)=1\}$. Trivially we have that $|(\mathbb{Z}/n\mathbb{Z})^{\times}|=\varphi(n)$. From there, there exists a sufficient $k$ such that $a^{k}\equiv 1\pmod{n}$, and by Lagrange's Theorem $bk|\varphi(n)$ which means $a^{\varphi(n)}\equiv a^{bk} \equiv (a^{k})^b\equiv 1^b\equiv 1\pmod{n}$, completing the proof. $\square$

Problems

Introductory

  • (BorealBear) Find the last two digits of $7^{81}-3^{81}$. (Solution)
  • (BorealBear) Find the last two digits of $3^{3^{3^{3}}}$. (Solution)
  • (2018 AMC 10B) Let $a_1,a_2,\dots,a_{2018}$ be a strictly increasing sequence of positive integers such that \[a_1+a_2+\cdots+a_{2018}=2018^{2018}.\]

What is the remainder when $a_1^3+a_2^3+\cdots+a_{2018}^3$ is divided by $6$?

\[\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4\]

  • (1983 AIME) Let $a_n=6^{n}+8^{n}$. Determine the remainder upon dividing $a_ {83}$ by $49$.

See also

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