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For a unit modulo n $a$ and a nonnegative integer $k$, division with remainder writes $k=q\phi (n)+r$ with $0\le r<\phi (n)$. Euler's theorem then givesThis replaces a potentially enormous exponent by a small remainder. The exponent itself is reduced modulo Euler's totient function, while the powers are reduced modulo $n$: these are different integer moduli.

For a worked modulo-fifteen calculation, seven is coprime to fifteen and $\phi (15)=8$. ThereforeUse exponentiation by squaring on the remaining exponent $5=(101{)}_2$, reducing after each multiplication:Thus $7^{2029}\mathrm{mod}15=7$. Modular exponentiation supplies the computation; Euler's theorem supplies the optional shortening of its exponent. Exponentiation by squaring also works for nonunit bases, without using this shortening.

The guaranteed exponent $\phi (n)$ is a common return exponent and an upper bound on the shortest period; it need not be that period. The multiplicative order $d={\mathrm{ord}}_n(a)$ is the least positive return exponent for the chosen unit. The Carmichael function $\lambda (n)$ is the least positive return exponent that works for every unit. Their precise relation isEither smaller return exponent can replace $\phi (n)$ in exponent reduction for units. Modulo fifteen, ${\mathrm{ord}}_{15}(7)=\lambda (15)=4<8=\phi (15)$, so $2029=507\cdot 4+1$ gives $7^{2029}\equiv 7$ immediately. For base four the multiplicative order is only two, smaller even than $\lambda (15)$. A return at exponent eight alone does not prove that the multiplicative order is eight.^[ref]

If the reduced exponent is zero, the rule gives $a^k\equiv a^0=1$, and still requires a unit. Applying it to the nonunit three would falsely predict $3^8\equiv 3^0=1(\mathrm{mod}15)$; the actual residue is six. For negative exponents, first interpret the power using a modular multiplicative inverse, as in modular exponentiation.

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 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.

Suppose $X^2\equiv -1(\mathrm{mod}p)$. Then $X$ is a unit modulo n for modulus $p$, and Fermat's little theorem impliesSince $p$ is an odd prime number, $1$ and $-1$ are distinct residue classes. Therefore $(p-1)/2$ is even, and $p\equiv 1(\mathrm{mod}4)$.

Conversely, put $q=(p-1)/2$. Pair each integer $j\in \{1,\dots ,q\}$ with $p-j$ in the factorial. Wilson theorem yieldsWhen $p\equiv 1(\mathrm{mod}4)$, $q$ is even, so $X=q!$ satisfies $X^2\equiv -1(\mathrm{mod}p)$. This proves the first supplementary law for quadratic reciprocity without assuming a primitive root. HenceThe two solutions are $q!$ and $-q!$ modulo $p$: their difference is nonzero, and factoring $X^2-Y^2$ shows that a degree-two polynomial over the field of residues modulo $p$ has no further roots.

Assume $ad\equiv bc(\mathrm{mod}n)$ and choose $r,s$ as in part (b). Put $\lambda =rc+sd$. Multiplying $ra+sb\equiv 1$ by $c$ and using $ad\equiv bc$ givesSimilarly, multiplication by $d$ gives $\lambda b\equiv d(\mathrm{mod}n)$.

Moreover, $\lambda$ is a unit modulo n: if a prime divided both $\lambda$ and $n$, the two congruences would make it divide $c,d,n$, contrary to $(c,d)\in Y$. Thus $R$ means precisely that two pairs differ by multiplication by a unit modulo $n$. Reflexivity, symmetry using ${\lambda }^{-1}$, and transitivity using products of units now show that $R$ is an equivalence relation.

For a positive modulus $n$, a reduced residue system contains exactly one representative of each residue class coprime to $n$. Its size is the Euler totient function $\phi (n)$. Multiplication by any unit modulo n permutes these classes, which proves the Fermat-Euler theorem after multiplying and cancelling their unit product.

For a positive integer modulus $m$, the residue class of an integer $a$ is the equivalence classIt contains all integers in the same position of the repeating cycle. For example,The residue class is this whole infinite collection; its canonical representative is the least nonnegative residue. Euclidean division gives exactly one such representative in $0,1,\dots ,m-1$, so there are exactly $m$ residue classes. When $m=1$, there is just one, $[0{]}_1=\mathbb{Z}$.

Define addition and multiplication of residue classes byThese definitions must give the same answer whichever integers represent the classes. If $a^{\prime }\equiv a$ and $b^{\prime }\equiv b$ modulo $m$, thenBoth differences are divisible by $m$, proving that the operations are well-defined. For example, $[9{]}_{12}+[5{]}_{12}=[2{]}_{12}$ and $[9{]}_{12}[5{]}_{12}=[9{]}_{12}$, since $45=3\cdot 12+9$. Replacing nine by $-3$ gives the same answers with smaller intermediate integers.

With these operations the residue classes form the finite commutative ring $\mathbb{Z}/m\mathbb{Z}$, a quotient ring. It need not be a field: $[6{]}_{12}[2{]}_{12}=[0{]}_{12}$ although neither factor is the zero class. These zero divisors explain why division requires a unit modulo n. Reducing after each multiplication is especially useful in modular exponentiation, where the unreduced powers can become enormous.

For a positive integer modulus $n$, a residue class $[a{]}_n$ is a unit when it has a modular multiplicative inverse: some integer $u$ satisfies $au\equiv 1(\mathrm{mod}n)$. Modulo twelve, the units are exactlyIn particular, five is its own modular inverse, since $5\cdot 5=25\equiv 1(\mathrm{mod}12)$. Multiplying all twelve residue classes by five gives a permutation; multiplying the outputs by five recovers the inputs. By contrast, multiplying by six gives only zero and six, with six inputs for each output. It cannot be reversed.

In general, the exact criterion isFor the forward direction, $au-kn=1$ implies every common divisor of $a,n$ divides one. Conversely, if their greatest common divisor is one, Bézout's identity supplies $ua+vn=1$, so $u$ is a modular inverse. The extended Euclidean algorithm computes these coefficients. The units form the unit group $(\mathbb{Z}/n\mathbb{Z}{)}^{\times }$ under multiplication. ^[ref]

Multiplicative cancellation while keeping the same integer modulus is valid for a unit modulo n: from $ca\equiv cb(\mathrm{mod}n)$, multiply by a modular inverse of $c$ to get $a\equiv b(\mathrm{mod}n)$. A nonunit cannot be cancelled as a general rule. For example, $6\cdot 0\equiv 6\cdot 2(\mathrm{mod}12)$, yet $0\not\equiv 2(\mathrm{mod}12)$. Thus division by six cannot undo this multiplication; solving a linear congruence instead keeps track of all possible classes.

This distinction marks the boundary between commutative rings and fields. Modulo a prime number $p>1$, every nonzero residue class is a unit, so $\mathbb{Z}/p\mathbb{Z}$ is a finite field. Modulo twelve it is a commutative ring with zero divisors, and is not a field. For $n=1$, the sole class satisfies $[0{]}_1=[1{]}_1$ and is a unit; this one-element zero ring is not a field. ^[ref]