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For an integer modulus $n\ge 2$ and an integer $a$ with greatest common divisor $\gcd (a,n)=1$, Euler's theorem, also called the Fermat-Euler theorem, statesHere Euler's totient function $\phi (n)$ counts the units in the unit group $U(n)=(\mathbb{Z}/n\mathbb{Z}{)}^{\times }$. These congruence classes have modular multiplicative inverses, equivalently their representatives are coprime to $n$. The guaranteed return to one permits exponent reduction for units.^[ref]

For a prime number $p$, $\phi (p)=p-1$, recovering Fermat's little theorem in the form $a^{p-1}\equiv 1(\mathrm{mod}p)$ when $p\nmid a$. Multiplying by $a$ and separately handling $p\mid a$ gives $a^p\equiv a(\mathrm{mod}p)$ for every integer $a$. For a composite number used as the integer modulus, a nonzero congruence class can be a nonunit.

Modulo twelve, the units have representatives $1,5,7,11$, so Euler's totient function gives $\phi (12)=4$. Multiplication by five producesThus this permutation has two permutation cycles, each of length two. In particular $5^2\equiv 1(\mathrm{mod}12)$, and the guaranteed return $5^4\equiv 1$ follows. The number of units need not be the length of one permutation cycle. Modulo fifteen, multiplication by seven similarly gives the two permutation cycles $(1,7,4,13)$ and $(2,14,8,11)$, covering all eight units. The powers of seven follow the permutation cycle containing one; the other permutation cycle shows that these powers do not visit every unit.

There is a stronger obstruction for a nonunit than just an occasional failed calculation. If a positive integer $d>1$ divides both $a$ and $n$, then $a^k\equiv 0(\mathrm{mod}d)$ for every $k\ge 1$, so $a^k$ cannot be congruent to one modulo $n$. For example, $\gcd (3,15)=3$ and $3^{\phi (15)}=3^8\equiv 6(\mathrm{mod}15)$, rather than one. Such a base has no multiplicative order modulo that integer modulus.^[ref]

Euler's totient function is the arithmetic functionIt counts the congruence classes modulo $n$ represented by integers coprime to $n$. In particular, $\phi (1)=1$, since $\gcd (1,1)=1$. For $12$, the coprime integers among $1,\dots ,12$ are $1,5,7,11$, so $\phi (12)=4$; they form a reduced residue system. ^[ref]

Counting these congruence classes is useful because they are exactly the classes with modular multiplicative inverses. Their number supplies the exponent in Euler theorem. Möbius inversion gives the product formulawhere $p$ runs over distinct prime factors of $n$. Thus $\phi (p^a)=p^{a-1}(p-1)$ for prime powers with $a\ge 1$, and $\phi (12)=12(1-1/2)(1-1/3)=4$. The product for $n=1$ is an empty product, so it gives the same convention $\phi (1)=1$.

Choose a reduced residue system $u_1,\dots ,u_{\phi (n)}$. For a unit modulo n $a$, the bijection $u\mapsto au$ on the unit group has inverse $u\mapsto a^{-1}u$. Comparing products before and after this permutation givesThe product $P$ is a unit; its modular inverse cancels it, proving Euler's theorem. Merely being nonzero would not justify cancellation.^[ref]

In the modulo-twelve example, $P=1\cdot 5\cdot 7\cdot 11=385\equiv 1(\mathrm{mod}12)$. The permuted representatives $5,1,11,7$ have the same product, while the unreduced products $(5\cdot 1)(5\cdot 5)(5\cdot 7)(5\cdot 11)$ equal $5^{4}P$. Equating these products modulo twelve exposes the fourth power without calculating it directly. Using all nonzero congruence classes instead would include nonunits, whose product need not permit cancellation.

The Fermat-Euler theorem states that, for a positive integer $q$ and an integer $a$ coprime to $q$, $a^{\phi (q)}\equiv 1(\mathrm{mod}q)$, where the Euler totient function $\phi (q)$ counts residue classes coprime to $q$. The case $q=1$ is trivial. For $q>1$, let $r_1,\dots ,r_{\phi (q)}$ be a reduced residue system. Multiplication by $a$ permutes it: equality of two resulting residues can be cancelled using the modular inverse of $a$, and the products remain coprime to $q$. Multiplying the residues before and after the permutation givesEvery factor on the right is a unit, so its product can be cancelled. This provesThe integer $359$ is prime: none of the primes $2,3,5,7,11,13,17$ up to $\sqrt{359}<19$ divides it. Therefore $\phi (359)=358$. Apply the theorem to $60$ and use the supplied congruence:To connect this to individual digits rather than merely assert periodicity, use long-division remainders $r_0=1$ andInduction gives $r_n\equiv 10^n(\mathrm{mod}359)$. Since $10$ is coprime to $359$, no remainder is zero. The congruence above therefore implies $r_{n+179}=r_n$ for every $n\ge 0$, using the unique representative in $1,\dots ,358$. In particularHenceThis is the mechanism of decimal period from multiplicative order: the period divides $179$.

The Fermat-Euler theorem states that for an integer $a$ and a positive integer $n$ with $\gcd (a,n)=1$,where the Euler totient function $\phi (n)$ counts the residue classes coprime to $n$. For $n\ge 2$, choose a reduced residue system $r_1,\dots ,r_{\phi (n)}$. Multiplication by the unit modulo n $a$ preserves coprimality and is injective on these residue classes: $a{r}_i\equiv a{r}_j(\mathrm{mod}n)$ implies $r_i\equiv r_j(\mathrm{mod}n)$, because $a$ has a multiplicative inverse modulo $n$. It therefore permutes the reduced residue system, givingThe product is also a unit modulo n, so it can be cancelled. This proves the Fermat-Euler theorem. For $n=1$ every modular congruence is automatic; one may use $\phi (1)=1$.

For a prime number $p$, the Euler totient function has value $\phi (p)=p-1$. Thus $a^{p-1}\equiv 1(\mathrm{mod}p)$ whenever $p\nmid a$. Multiplying by $a$, and treating $p\mid a$ separately, gives the all-integer version of Fermat's little theorem:Finally, Wilson theorem states that for every prime number $p$, $\boxed{(p-1)!\equiv -1(\mathrm{mod}p)}$. Its converse is also true: an integer $n>1$ is prime exactly when $(n-1)!\equiv -1(\mathrm{mod}n)$. For the converse, any proper divisor $1<d<n$ would divide both $(n-1)!$ and $n$, contradicting that congruence.