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For the period matrix of a genus-$g$ compact Riemann surface, there is a vector $\kappa$ depending on the marking and base point such that the Riemann theta function vanishes exactly on this translate of the image of the Abelian sum map. Pairing sides of the cut surface proves that a nonzero pullback $\theta (u(p)-z)$ has a degree-$g$ zero divisor whose Abel sum is $z+\kappa$. At a theta zero that pullback includes the base point. Irreducibility and dimension of the effective-divisor locus give equality of supports; the primitive First Chern class of the theta line bundle makes the divisor reduced.

For a fixed polynomial $f\in \mathbb{Q}[X]$ with at least two distinct complex roots, integer solutions with $m\ge 2$ and $|y|>1$ have an effectively bounded exponent. Factorization over a splitting field and the fixed differences between distinct roots control common ideal divisors. Outside a fixed set of primes, each factor's valuation is divisible by $m/\gcd (m,e_i)$, where $e_i$ is its root multiplicity. The generalized power-equation estimates from linear forms in logarithms of algebraic numbers, including unit and exceptional-prime factors, give the exponent bound. A single-root polynomial and the values $y=0,\pm 1$ cannot be included: $4^m=(2^m{)}^2$, and $1^m=f(0)$ for $f(X)=X(X-1)+1$, give infinite exponent families.

For real coefficients ${\rho }_1=1$ supported on permitted divisors up to $L$, put ${\lambda }_d={\sum }_{[u,v]=d}{\rho }_u{\rho }_v$. Then ${\sum }_{d\mid n}{\lambda }_d=({\sum }_{d\mid n}{\rho }_d{)}^2$ is an upper bound for the indicator function of numbers free of the forbidden prime factors. Its support has $d\le L^2$. If the coefficients vanish on non-squarefree integers and satisfy $|{\rho }_d|\le 1$, then $|{\lambda }_d|\le 3^{\omega (d)}$: each prime factor of $d$ occurs in $u$, in $v$, or in both.

A squarefree divisor of an integer $n$ is a divisor $d\mid n$ that is a squarefree integer. It is specified by choosing a subset of the distinct prime factors of $n$.

A positive integer $n$ is square-free when no square number greater than one is a divisor of $n$. It suffices to exclude squares of prime numbers; equivalently, every prime valuation is zero or one:Thus the prime factorization of a square-free integer is a product of distinct prime factors, with no repetitions. One is included: its prime factorization is an empty product and all its prime valuations are zero. These equivalent tests are stated in Grinberg's notes. ^[ref]

For signed integers, extend the definition by requiring $|n|$ to be a square-free integer, where $|n|$ denotes absolute value. Hence $-30$ and both $1,-1$ qualify. Zero does not: every prime number has its square as a divisor of zero. The Möbius function and radical of an integer below use their existing positive integer domains, so apply them to $|n|$ when handling a negative integer.

For example, $30=2\cdot 3\cdot 5$ is a square-free integer, whereas $12=2^2\cdot 3$ is not, because $4\mid 12$ and $v_2(12)=2$. Being a composite number causes no problem: $6,10,15$ are all square-free integers. A square-free integer need not be coprime to another integer: $6$ and $10$ are both square-free integers but have greatest common divisor two. In fact the only signed integers coprime to every integer are $1,-1$, since $\gcd (n,n)=|n|$.

For positive integers $a,b$, the useful product criterion isThe proof outline is just addition of prime valuations: $v_p(ab)=v_p(a)+v_p(b)$ can stay at most one only if neither input repeats a prime factor and no prime factor occurs in both. Thus $6\cdot 5=30$ is a square-free integer, but $6\cdot 10=60$ is not. An equivalent characterization is that every factorization $n=ab$ into positive integers has coprime factors. If $p^2\mid n$, the factorization $n=p(n/p)$ supplies two factors sharing $p$; conversely, a shared prime factor forces its square to divide $n$. For $30$, the unordered factor pairs $(1,30),(2,15),(3,10),(5,6)$ all have greatest common divisor one; for $12$, the pair $(2,6)$ fails.

Two existing arithmetic functions turn the definition into practical tests:The radical of an integer keeps one copy of each prime factor, so it changes precisely the integers with a repeated prime factor. Its definition is given in Evertse's exercise sheet. ^[ref] The Möbius function assigns $(-1{)}^r$ to a square-free integer with $r$ distinct prime factors and zero otherwise; squaring forgets the sign. Consequently ${\sum }_{n=1}^{60}\mu (n{)}^2=37$ counts exactly the green cells. This connects a restriction on repeated prime factors with weighted counting and inclusion-exclusion. ^[ref]

If $p\mapsto \theta (u(p)-z)$ is not identically zero, its zero divisor $D_z$ has these properties. The logarithmic derivative changes by $-2\pi i{\omega }_j$ across the paired $b_j$ translation, giving zero count $g$. For a variation of $z$, the logarithm changes there by $2\pi i\delta z_j$; integrating by parts in the first-moment contour integral gives $\delta \mathrm{AJ}(D_z)=\delta z$. Thus the difference is constant. Weighted zero divisors, rather than a distinct-zero assumption, make the statement valid at collisions.

Trial division searches for divisors of an integer by testing candidate divisions. For primality testing of an integer $n>1$, testing the prime numbers up to $\sqrt{n}$ suffices: any composite number has a prime factor in that range. After finding a prime factor, repeated division removes its prime power, and testing the remaining quotient gives an elementary integer factorization method. This is useful for small inputs, but the number of candidates can grow rapidly with the input's digit count.

Factor a positive integer while multiplying the exponent-plus-one contributions to its divisor count. After removing all prime factors below the current trial factor, a trial factor that divides the remainder must itself be prime. Remove its full power and advance. Once the trial factor's square exceeds the remainder, any remainder greater than one is prime and contributes a factor two. The invariant separates the original integer into removed prime powers and the remaining coprime factor; no factor array is necessary.

A divisor $d$ of a positive integer $n$ is unitary when $\gcd (d,n/d)=1$. Each prime power in the prime factorization of $n$ must therefore be included in $d$ completely or omitted completely. Consequently, the sum of the $k$-th powers of unitary divisors is ${\prod }_{p^e\parallel n}(1+p^{ek})$.

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]

The displayed identity is exact for all positive integers and all positive cutoffs. Expand $(\mu -{\mu }_{\le U})\ast (\mu -{\mu }_{\le V})\ast 1$ and use $\mu \ast 1=\epsilon$. The two truncated terms are short sums; the negative term is a type I sum over divisors at most $UV$; the last term is a type II sum with both factor variables larger than a cutoff. This form of the Vaughan identity is specific to Möbius weights.

For a period lattice, the displayed canonical product converges normally because its logarithmic tails are bounded by a constant times $|\omega {|}^{-3}$. It defines an odd entire function with simple zeros exactly at the lattice and derivative one at zero. Its logarithmic derivative is the Weierstrass zeta function, and the derivative of that logarithmic derivative is minus the Weierstrass elliptic function. Sigma is quasi-periodic rather than elliptic; balanced products of its translates construct meromorphic functions with prescribed divisors on a complex torus.

For an integer $n>1$, Wilson's theorem characterizes prime numbers by a factorial:The right side is an integer congruence, so the least nonnegative remainder is $n-1$. For example, $4!=24\equiv 4(\mathrm{mod}5)$ certifies that five is a prime number, whereas $5!=120\equiv 0(\mathrm{mod}6)$ rules out six. ^[ref]

For an odd prime number $p$, each nonzero congruence class has a unique modular multiplicative inverse. Pair each class with its inverse: each distinct pair contributes one. A class is its own inverse exactly when $a^2\equiv 1(\mathrm{mod}p)$; since $p\mid (a-1)(a+1)$, Euclid's lemma leaves only $a\equiv \pm 1$. Their product is $-1$. The prime number two needs a separate check: $1!\equiv -1(\mathrm{mod}2)$. Conversely, a proper divisor $1<d<n$ divides both $n$ and $(n-1)!$, so the displayed integer congruence would force $d\mid 1$, a contradiction. ^[ref]

The composite numbers have an even sharper pattern: $(n-1)!\equiv 0(\mathrm{mod}n)$ for every composite number $n>4$, but $3!=6\equiv 2(\mathrm{mod}4)$. To check the zero claim, write $n=ab$ with $1<a\le b<n$. If $a<b$, both occur as different factors in the factorial. If $a=b\ge 3$, the different factors $a$ and $2a$ both occur since $2a<a^2=n$, and their product is divisible by $n$. The only excluded case is $a=b=2$. Thus four is an exception to the zero pattern, but still fails Wilson's theorem: its required remainder would be three.

This is an exact primality testing criterion, but direct evaluation is expensive: even reducing modulo $n$ after every multiplication still takes a number of multiplications proportional to $n$, which grows exponentially with the number of binary digits of $n$. The problem is not merely storing a huge factorial. The more useful role of Wilson's theorem is simplifying factorial expressions in modular arithmetic. For instance, if $p>2$ is a prime number, then $(p-1)(p-2)!\equiv -1(\mathrm{mod}p)$ gives $(p-2)!\equiv 1(\mathrm{mod}p)$ by a modular multiplicative inverse.