Difference between revisions of "Euler's Totient Theorem"

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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 known as Euler's generalization and Fermat-Euler as well.
 
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 known as Euler's generalization and Fermat-Euler as well.
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==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. 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 \cdot 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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== See also ==
 
== See also ==

Revision as of 20:28, 6 May 2013

Euler's Totient Theorem is a theorem closely related to his function of the same name.

Theorem

Let $\phi(n)$ be Euler's totient function. 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 that ${m}$ is prime. For this reason it is known as Euler's generalization and Fermat-Euler as well.

Proof

Consider the set of numbers $A =${$n_1, n_2, ... n_{\phi(m)}$} (mod m) such that the elements of the set are the numbers relatively prime to each other. It will now be proved that this set is the same as the set $B =${$an_1, an_2, ... an_{\phi(m)}$} (mod 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 $A$. This means that $n_1 \cdot n_2 ... n_{\phi(m)} \equiv an_1 \cdot an_2 ... an_{\phi(m)}$(mod m) => $a^{\phi (m)} \cdot (n_1 n_2 ... n_{\phi(m)}) \equiv n_1 n_2 ... n_{\phi(m)}$ (mod m) => $a^{\phi (m)} \equiv 1$ (mod m) as desired.


See also