Difference between revisions of "Euler's Totient Theorem"

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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>.  
 
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>.  
 
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> (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)}</math><math>\pmod{m}</math> → <math>a^{\phi (m)} \cdot (n_1 n_2 ... n_{\phi(m)}) \equiv n_1 n_2 ... n_{\phi(m)}</math><math>\pmod{m}</math> → <math>a^{\phi (m)} \equiv 1</math><math>\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>.
 
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> (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)}</math><math>\pmod{m}</math> → <math>a^{\phi (m)} \cdot (n_1 n_2 ... n_{\phi(m)}) \equiv n_1 n_2 ... n_{\phi(m)}</math><math>\pmod{m}</math> → <math>a^{\phi (m)} \equiv 1</math><math>\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>.
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==Problems==
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===Introductory===
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*(BorealBear) Find the last two digits of <math> 5^{81}-3^{81} </math>.
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*(BorealBear) Find the last two digits of <math> 3^{3^{3^{3}}} </math>.
  
 
== See also ==
 
== See also ==

Revision as of 16:10, 22 April 2021

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.

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 $(a, m) = 1$. 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)}$$\pmod{m}$$a^{\phi (m)} \cdot (n_1 n_2 ... n_{\phi(m)}) \equiv n_1 n_2 ... n_{\phi(m)}$$\pmod{m}$$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 $m$.

Problems

Introductory

  • (BorealBear) Find the last two digits of $5^{81}-3^{81}$.
  • (BorealBear) Find the last two digits of $3^{3^{3^{3}}}$.

See also