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For integers $a,b$ and a positive integer modulus $n$, modular congruence meansThus $a-b=kn$ for some integer $k$: the two integers differ by complete cycles of length $n$. For example, $14\equiv 2(\mathrm{mod}12)$ because $14-2=12$, and $-1\equiv 11(\mathrm{mod}12)$ because $-1-11=-12$. A modular congruence need not be an equality of integers.

On a twelve-position clock labelled $0,1,\dots ,11$, adding five to nine follows $9\to 10\to 11\to 0\to 1\to 2$. Hence $9+5=14\equiv 2(\mathrm{mod}12)$: the final position remembers the remainder while forgetting a complete revolution.

The same idea tracks any count repeating every $n$ steps. It also lets us compute sums and products with small residue classes, reducing intermediate results without changing the final class. The justification comes from addition and multiplication being well-defined on residue classes.

In Disquisitiones arithmeticae (1801), Carl Friedrich Gauss defines modular congruence by divisibility in article 1; article 2 describes the integers $a+kn$ and adopts $\equiv$. These passages document his presentation and notation, without claiming that repeating remainders began in 1801. ^[ref]

For a prime number $p$ and $p\nmid a$, multiplication by $a$ gives a permutation of the $p-1$ nonzero residue classes. For $a=3$ modulo seven, the representatives $1,2,3,4,5,6$ map to $3,6,2,5,1,4$: the same collection in a different order. Repeating this permutation from one visits all six representatives before returning.

For $a=2$ modulo seven, the permutation cycles are instead $(1,2,4)$ and $(3,6,5)$. The powers of two visit only the permutation cycle containing one, so their multiplicative order is three. Both permutation cycles return after three steps; applying the permutation six times still fixes every nonzero residue class. Fermat's little theorem guarantees a return at six, not that every base visits every nonzero residue class.

More generally, every permutation cycle of this map has length equal to the multiplicative order of $a$: starting at a nonzero residue class $x$, a return means $a^{k}x\equiv x(\mathrm{mod}p)$, and its modular multiplicative inverse allows cancellation of $x$. Thus the $p-1$ nonzero residue classes split into equal-length permutation cycles, which also explains the divisibility of the multiplicative order into $p-1$.

The product argument gives a short proof outline. Since multiplication by $a$ is a permutation of the nonzero residue classes,Every factor is a unit, so their product has a modular multiplicative inverse and can be cancelled. This leaves $a^{p-1}\equiv 1(\mathrm{mod}p)$. The whole argument works for $p=2$, where the product is just one.^[ref]

A rational homogeneous equation ${\sum }_i{a}_i{x}_i=0$ has a monochromatic positive odd solution in every finite colouring of the odd integers exactly when its full coefficient sum is zero. Sufficiency uses a constant odd vector. For necessity, clear denominators and, if the total $S$ is nonzero, colour by odd residue classes modulo $2^K>|S|$. A common odd residue is invertible, so a solution would require $2^K\mid S$, impossible. This is stronger than the zero-sum-subset criterion for unrestricted partition regularity.

For a prime number $p$, the nonzero residue classes form the multiplicative group of the finite field ${\mathbb{F}}_p$. Every class has a unique multiplicative inverse. A class equals its own inverse precisely when $x^2=1$, so $(x-1)(x+1)=0$ in the field and $x=1$ or $x=-1$.

For an odd prime, pair each remaining class with its distinct inverse. Each pair contributes one to the product, leavingFor $p=2$, the same congruence is $1\equiv -1(\mathrm{mod}2)$. This proves Wilson's theorem.

If $p$ is a prime number and $p\nmid a$, multiplication by $a$ permutes the nonzero residue classes modulo $p$: $ai\equiv aj$ implies $i\equiv j$ because $a$ has a multiplicative inverse. Multiplying all the permuted classes givesThe factorial is a nonzero element of the finite field ${\mathbb{F}}_p$, so cancellation is valid. Thereforewhich is Fermat's little theorem.

Fix a positive integer $q$ and an integer $a$ coprime to $q$. We use Dirichlet characters modulo $q$, extended by zero on integers not coprime to $q$, and write ${\chi }_0$ for the principal Dirichlet character. For a nonprincipal Dirichlet character, the sum over one complete period is zero. Its partial sums $A_{\chi }(u)={\sum }_{n\le u}\chi (n)$ are therefore $O(q)$. Partial summation givesThe integral converges locally uniformly there, so these Dirichlet L-functions are holomorphic at one. On the other hand,has a simple pole at one, with residue $\phi (q)/q$.

The assumed nonvanishing of a nonprincipal Dirichlet L-function at one for real Dirichlet characters also excludes vanishing for nonreal ones. To prove this, form the product over all Dirichlet charactersUse the logarithm supplied by the absolutely convergent Euler product. The Orthogonality of Dirichlet characters yieldsThus $\mathcal{F}(\sigma )\ge 1$. If a nonreal Dirichlet character $\chi$ had $L(1,\chi )=0$, its distinct conjugate $\overline{\chi }$ would also have a zero there, since $L(1,\overline{\chi })=\overline{L(1,\chi )}$. These two zeros would outweigh the single pole of the principal factor; all other factors are bounded near one. It would follow that $\mathcal{F}(\sigma )\to 0$ as $\sigma \downarrow 1$, contradicting the lower bound. Together with the assumption for real Dirichlet characters, this proves $L(1,\chi )\ne 0$ for every nonprincipal Dirichlet character.

The first-degree part of the Euler product logarithm givesThe omitted terms are bounded uniformly, since ${\sum }_p{\sum }_{m\ge 2}(m{p}^{m\sigma }{)}^{-1}\ll {\sum }_p{p}^{-2}<\infty$. For each nonprincipal factor, its logarithm remains bounded: choose a holomorphic logarithm near the now established nonzero value $L(1,\chi )$; the Euler-product branch differs from it by a fixed integral multiple of $2\pi i$ on a sufficiently small real interval. For the principal factor,Applying Orthogonality of Dirichlet characters again givesThe right side diverges as $\sigma \downarrow 1$, so the arithmetic progression contains infinitely many primes. This proves Dirichlet's theorem on arithmetic progressions: every reduced residue class modulo $q$ contains infinitely many primes. The argument also covers $q=1$ and $q=2$, when there need not be a nonreal character.

For the sharper statement about reciprocal primes in a fixed arithmetic progression, use the given prime-counting function asymptotic with $q$ fixed:The second equivalence follows by integration by parts in the logarithmic integral function. Partial summation yieldsFor any $\epsilon >0$, choose a fixed $T$ so large that the relative error in $\pi (t;q,a)$ is at most $\epsilon$ for $t\ge T$. The initial integral is a constant, while the remaining integral lies betweenThe boundary term is $O_q(1/\log x)$. Divide by $\log \log x$, let $x\to \infty$, and then let $\epsilon \downarrow 0$. We obtainOnly a relative asymptotic is deduced from the given information; an additive $O(1)$ error, or uniformity for a growing modulus $q$, would require additional estimates.

An integer square is divisible by four exactly when its base is even: an even base has square $4j^2$, while an odd base has square $(2j+1{)}^2\equiv 1(\mathrm{mod}4)$. Apply this observation to the triangular number $T_n$ obtained above. Since $n(n+1)$ is even,Checking the four residue classes of $n$ modulo four, the products $n(n+1)$ have residues $0,2,2,0$, respectively. ThereforeThe original PDF has two alternatives modulo four; the TeX transcription incorrectly turns the second alternative into a modulus-three expression.

The modular congruence $x^4\equiv 1$ first implies $x\not\equiv 0$. Multiplication by such an $x$ permutes the nonzero residue classes modulo the prime $p$: if $xa\equiv xb$, cancellation gives $a\equiv b$. Multiply all these residues and cancel their nonzero product to obtain $x^{p-1}\equiv 1$, the usual short proof of Fermat little theorem.

Now $(x^2-1)(x^2+1)\equiv 0$. By the Euclid lemma, either $x^2\equiv 1$ or $x^2\equiv -1$. In the latter case, because $p-1=4k+2$,contradicting the preceding product argument, since $p$ is odd. ThusNo unproved assertion about the number of fourth roots is needed.

For a prime number $p$, every nonzero residue class has a unique multiplicative inverse modulo $p$: Bezout identity supplies one because its representative is coprime to $p$, and cancellation modulo a prime gives uniqueness. Pair each residue with its inverse. The only unpaired residues satisfy $a^2\equiv 1$, so $p\mid (a-1)(a+1)$; the Euclid lemma gives $a\equiv 1$ or $-1$. For odd $p$, these are distinct. Every other pair contributes one to the product of all nonzero residues, leavingFor $p=2$, $1!\equiv -1(\mathrm{mod}2)$ as well. This proves Wilson theorem in all cases.

For an odd prime, set $q=(p-1)/2$. Pair the upper half of the factorial with the negatives of the lower half:If $p\equiv 1(\mathrm{mod}4)$, then $q$ is even. ThereforeThis constructs a square root of minus one modulo a prime explicitly.

Fix a modulus $q$ and a residue class $a$ with $(a,q)=1$. The Dirichlet characters modulo $q$ are the one-dimensional characters of a representation of $(\mathbb{Z}/q\mathbb{Z}{)}^{\times }$, extended by zero on nonunits. Let ${\chi }_0$ be the principal Dirichlet character.

For every nonprincipal Dirichlet character the partial sums ${\sum }_{n\le u}\chi (n)$ are bounded: the sum over a full period is zero, and a remaining incomplete period has bounded length. Partial summation therefore continues its Dirichlet L-function holomorphically to $\Re s>0$ byThis shows in particular that it has no pole at one. By contrasthas a simple pole there. We use the stipulated nonvanishing $L(1,\chi )\ne 0$ for every real nonprincipal Dirichlet character, and establish the remaining nonvanishing as follows.

For real $\sigma >1$, take the product over all Dirichlet characters. The Euler product logarithms and Orthogonality of Dirichlet characters giveThus this product is real and at least one. If a nonreal Dirichlet character had $L(1,\chi )=0$, complex conjugation would also give $L(1,\overline{\chi })=0$. These are distinct factors. Two zeros and only one possible simple pole force the whole product to tend to zero as $\sigma \downarrow 1$, contradicting the lower bound. Hence all nonprincipal Dirichlet characters have $L(1,\chi )\ne 0$, under the given real-character assumption.

For $\sigma >1$, absolute convergence of each Euler product giveswith bounded remainder as $\sigma \downarrow 1$: the terms of prime-power exponent at least two have total modulus at most ${\sum }_{n\ge 2}1/(n(n-1))$. Since a nonprincipal $L$ defines a holomorphic function that is nonzero near one, its Euler-product logarithm differs from a local holomorphic logarithm by a fixed integer multiple of $2\pi i$ on a short real interval. It is therefore bounded there. For the principal Dirichlet character the pole instead givesApply the Dirichlet character orthogonality relation at each prime. The prime-character sum near one yieldsThis tends to infinity. A finite set of primes would have a bounded sum as $\sigma \downarrow 1$, so every coprime residue class contains infinitely many primes. This proves the Dirichlet theorem on primes in arithmetic progressions from the specified assumption. In fact it proves divergence of the corresponding reciprocal-prime series and gives Dirichlet density $1/\phi (q)$.

Finally, for real $s>1$ each prime satisfiesThe integrands are nonnegative, so Tonelli theorem permits summation inside the integral. At a fixed $t$, exactly $\pi (t,q,a)$ primes from the residue class have entered. ThereforeBoth sides are finite, since they are bounded respectively by ${\sum }_{n\ge 2}{n}^{-s}$ and $s{\int }_1^{\infty }{t}^{-s}dt$. The values of a counting function at its isolated jump endpoints do not change this integral.

The Von Mangoldt function is supported on prime powers:Write $e(t)=\exp (2\pi it)$. A convenient form of the Siegel–Walfisz theorem is: for every fixed $A,B>0$, uniformly for $q\le (\log x{)}^B$ and $(a,q)=1$,The implied constant can be ineffective because of possible Siegel zeros; the modulus bound is logarithmic, not an unrestricted uniformity assertion.

Apply this to modulus five and separate the reduced residue classes:The last sum is $O(\log N)$: only powers of five can contribute in the zero residue class. By the orthogonality of roots of unity,since two is coprime to five and the sum including $a=0$ vanishes. As $\phi (5)=4$, the Siegel–Walfisz theorem yieldsThe sum is complex-valued, but the displayed error controls its absolute difference from the real main term.