Euler's theorem (number theory)

Euler's theorem in number theory says:

If a $a$ ${\ displaystyle a}$ $a$ and $m$ ${\ displaystyle m}$ $m$ mutually simple then $a^{\varphi (m)}\equiv 1{\pmod {m}}$ ${\ displaystyle a ^ {\ varphi (m)} \ equiv 1 {\ pmod {m}}}$ $a ^ {\ varphi (m)} \ equiv 1 \ pmod m$ where $\varphi (m)$ ${\ displaystyle \ varphi (m)}$ $\ varphi (m)$ - Euler function .

An important consequence of the Euler theorem for the case of a simple module is the small Fermat theorem :

If a $a$ ${\ displaystyle a}$ $a$ not divisible by prime $p$ ${\ displaystyle p}$ $p$ then $a^{p-1}\equiv 1{\pmod {p}}$ ${\ displaystyle a ^ {p-1} \ equiv 1 {\ pmod {p}}}$ $a ^ {{p-1}} \ equiv 1 {\ pmod p}$ .

In turn, the Euler theorem is a consequence of the Lagrange general algebraic theorem applied to the reduced residue system modulo $m$ ${\ displaystyle m}$ $m$ .

Content

Evidence

Using number theory

Let be $x_{1},\dots ,x_{\varphi (m)}$ ${\ displaystyle x_ {1}, \ dots, x _ {\ varphi (m)}}$ - all different natural numbers, smaller $m$ ${\ displaystyle m}$ and mutually simple with it.

Consider all possible works. $x_{i}a$ ${\ displaystyle x_ {i} a}$ for all $i$ ${\ displaystyle i}$ from $one$ ${\ displaystyle 1}$ before $\varphi (m)$ ${\ displaystyle \ varphi (m)}$ .

Insofar as $a$ ${\ displaystyle a}$ mutually easy with $m$ ${\ displaystyle m}$ and $x_{i}$ ${\ displaystyle x_ {i}}$ mutually easy with $m$ ${\ displaystyle m}$ , then $x_{i}a$ ${\ displaystyle x_ {i} a}$ also mutually easy with $m$ ${\ displaystyle m}$ , i.e $x_{i}a\equiv x_{j}{\pmod {m}}$ ${\ displaystyle x_ {i} a \ equiv x_ {j} {\ pmod {m}}}$ for some $j$ ${\ displaystyle j}$ .

Note that all residues $x_{i}a$ ${\ displaystyle x_ {i} a}$ when divided by $m$ ${\ displaystyle m}$ are different. Indeed, let it not be so, then there are such $i_{1}\neq i_{2}$ ${\ displaystyle i_ {1} \ neq i_ {2}}$ , what

x_{i_{1}}a\equiv x_{i_{2}}a{\pmod {m}}

{\ displaystyle x_ {i_ {1}} a \ equiv x_ {i_ {2}} a {\ pmod {m}}}

or

(x_{i_{1}}-x_{i_{2}})a\equiv 0{\pmod {m}}.

{\ displaystyle (x_ {i_ {1}} - x_ {i_ {2}}) a \ equiv 0 {\ pmod {m}}.}

Because $a$ ${\ displaystyle a}$ mutually easy with $m$ ${\ displaystyle m}$ then the last equality is equivalent to

x_{i_{1}}-x_{i_{2}}\equiv 0{\pmod {m}}

{\ displaystyle x_ {i_ {1}} - x_ {i_ {2}} \ equiv 0 {\ pmod {m}}}

or

x_{i_{1}}\equiv x_{i_{2}}{\pmod {m}}

{\ displaystyle x_ {i_ {1}} \ equiv x_ {i_ {2}} {\ pmod {m}}}

.

This is contrary to the fact that the numbers $x_{1},\dots ,x_{\varphi (m)}$ ${\ displaystyle x_ {1}, \ dots, x _ {\ varphi (m)}}$ pairs are distinct in modulus $m$ ${\ displaystyle m}$ .

Multiply all comparisons of the form. $x_{i}a\equiv x_{j}{\pmod {m}}$ ${\ displaystyle x_ {i} a \ equiv x_ {j} {\ pmod {m}}}$ . We get:

x_{1}\cdots x_{\varphi (m)}a^{\varphi (m)}\equiv x_{1}\cdots x_{\varphi (m)}{\pmod {m}}

{\ displaystyle x_ {1} \ cdots x _ {\ varphi (m)} a ^ {\ varphi (m)} \ equiv x_ {1} \ cdots x _ {\ varphi (m)} {\ pmod {m}}}

or

x_{1}\cdots x_{\varphi (m)}(a^{\varphi (m)}-1)\equiv 0{\pmod {m}}

{\ displaystyle x_ {1} \ cdots x _ {\ varphi (m)} (a ^ {\ varphi (m)} - 1) \ equiv 0 {\ pmod {m}}}

.

Since the number $x_{1}\cdots x_{\varphi (m)}$ ${\ displaystyle x_ {1} \ cdots x _ {\ varphi (m)}}$ mutually easy with $m$ ${\ displaystyle m}$ the last comparison is equivalent to

a^{\varphi (m)}-1\equiv 0{\pmod {m}}

{\ displaystyle a ^ {\ varphi (m)} - 1 \ equiv 0 {\ pmod {m}}}

or

a^{\varphi (m)}\equiv 1{\pmod {m}}.

{\ displaystyle a ^ {\ varphi (m)} \ equiv 1 {\ pmod {m}}.}

■

Using group theory

Consider the multiplicative group $\mathbb {Z} _{n}^{*}$ {\ displaystyle \ mathbb {Z} _ {n} ^ {*}} reversible elements of the residue ring $\mathbb {Z} _{n}$ ${\ displaystyle \ mathbb {Z} _ {n}}$ . Its order is equal to $\varphi (n)$ ${\ displaystyle \ varphi (n)}$ according to the definition of the Euler function . Since the number $a$ ${\ displaystyle a}$ mutually easy with $n$ ${\ displaystyle n}$ corresponding element ${\overline {a}}$ ${\ displaystyle {\ overline {a}}}$ at $\mathbb {Z} _{n}$ ${\ displaystyle \ mathbb {Z} _ {n}}$ is reversible and belongs $\mathbb {Z} _{n}^{*}$ ${\ displaystyle \ mathbb {Z} _ {n} ^ {*}}$ . Element ${\overline {a}}\in \mathbb {Z} _{n}^{*}$ ${\ displaystyle {\ overline {a}} \ in \ mathbb {Z} _ {n} ^ {*}}$ generates a cyclic subgroup whose order, according to the Lagrange theorem , divides $\varphi (n)$ ${\ displaystyle \ varphi (n)}$ from here ${\overline {a}}^{\varphi (n)}={\overline {1}}$ ${\ displaystyle {\ overline {a}} ^ {\ varphi (n)} = {\ overline {1}}}$ . ■

Literature

Aierland K., Rosen M. A classic introduction to modern number theory. - M .: Peace, 1987.

Links

Topics: Euler's Theorem, Chinese Remainder Theorem , Amir Kamil, CS70, Fall 2003. UC Berkeley (Eng.)
RSA and Wagner, CS70, Fall 2003. UC Berkeley (English)