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A Pythagorean triple consists of three positive integers satisfying $x^2+y^2=z^2$. It is primitive when their greatest common divisor is one. Reducing their square numbers modulo $3$ shows that at least one leg is divisible by $3$; reducing modulo $4$ shows that both legs cannot be odd.

The radical of an integer $n>0$ is the product of its distinct prime factors, each included exactly once. If the prime factorization is $n=p_1^{a_1}\cdots p_r^{a_r}$ with distinct prime numbers $p_i$ and positive integer exponents $a_i$, thenThus the radical of an integer remembers which prime factors occur and forgets their prime valuations. At $n=1$, the empty product gives $\mathrm{rad}(1)=1$. The domain of a function here is the positive integers; zero is excluded. ^[ref]

For example, $72=2^3\cdot 3^2$ has distinct prime factors $2,3$, so $\mathrm{rad}(72)=2\cdot 3=6$. Similarly,using the prime factorizations $30=2\cdot 3\cdot 5$, $32=2^5$ and $36=2^2\cdot 3^2$. The prime omega function counts the retained prime factors, while the total number of prime factors counts their occurrences before repetitions are removed.

The radical of an integer is its largest squarefree divisor. Indeed, a squarefree divisor chooses at most one copy of each prime factor, so it divides $\mathrm{rad}(n)$; the radical of an integer itself is a squarefree divisor. Consequently,The last statement follows because every nonzero prime valuation is then one. It includes $n=1$, a squarefree integer. Applying the radical of an integer twice also changes nothing: $\mathrm{rad}(\mathrm{rad}(n))=\mathrm{rad}(n)$.

For any positive integers $a,b$, the exact product rule isTo prove it, a prime factor dividing just one input occurs once in the numerator and not in the denominator. A prime factor dividing both occurs twice in the numerator and once in the denominator, since it divides their greatest common divisor. Each prime factor of $ab$ therefore remains exactly once. For coprime integers, the denominator is $\mathrm{rad}(1)=1$, proving the radical of an integer is a multiplicative function. It is not a completely multiplicative function: $\mathrm{rad}(4)=2\ne 4=\mathrm{rad}(2{)}^2$.

For a worked example with shared prime factors, take $a=12$ and $b=18$. Their greatest common divisor is $6$, and $ab=216=2^3\cdot 3^3$, soThis explains why prime support of an integer is useful: questions about the presence of prime factors can discard their multiplicities. For example, the squarefree divisors of $72$ are exactly the integer divisors of $\mathrm{rad}(72)=6$, namely $1,2,3,6$. Also $\gcd (m,n)=1$ if and only if $\gcd (m,\mathrm{rad}(n))=1$, because coprimality depends only on shared prime factors.

A further characterization uses mutual divisibility of powers: for positive integers $a,b$,If their prime supports of an integer agree, sufficiently large $k,\ell$ make all the required prime valuations large enough. Conversely, these two divisibility relations force each prime factor of either input to divide the other. For $72$ and $36$, one can take $k=2$ and $\ell =1$: $72\mid 36^2$ and $36\mid 72$.

Exact arithmetic on rational numbers represented by integer numerators and nonzero integer denominators. Common factors can be removed with the greatest common divisor. Exact rational arithmetic avoids rounding error when deciding equality or integrality.

For a nonzero polynomial $g\in \mathbb{Q}[t]$, choose a positive integer $n$ clearing its denominators and set $c(g)=c(ng)/n$, where the polynomial content of an integer-coefficient polynomial is the positive greatest common divisor of its coefficients. This is independent of $n$: if $f/n=h/m$, then $mf=nh$, hence $mc(f)=nc(h)$.

The integer-coefficient product formula from Gauss lemma for polynomials givesMoreover $g/c(g)$ is an integer-coefficient primitive polynomial: if $g=f/n$, then $g/c(g)=f/c(f)$. For example,This normalization proves a useful divisibility descent. If $f\in \mathbb{Z}[t]$ is primitive, $g\in \mathbb{Z}[t]$ and $g=fq$ for $q\in \mathbb{Q}[t]$, then $c(q)=c(g)$ is an integer. Since $q/c(q)\in \mathbb{Z}[t]$, it follows that $q\in \mathbb{Z}[t]$.

Repeatedly apply the public RSA cryptosystem encryption map to a ciphertext. A return to that ciphertext reveals its preceding iterate as the plaintext. A nontrivial greatest common divisor between an iterate-minus-start and the modulus instead enables integer factorization. Eventual return alone supplies no useful running-time bound.

For positive integers $m,n,k$, the greatest common divisor scales by the common multiplier $k$. If $d=\gcd (m,n)$, then $kd$ is a common divisor of $km,kn$. Meanwhile Bézout's identity expresses $kd$ as an integer linear combination of $km,kn$, so their greatest common divisor divides $kd$. These two comparisons prove the identity without prime factorization.

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]

For nonzero coprime integers $p,q$, the torus knot $T_{p,q}$ lies on a standard unknotted torus in the three-sphere and winds $p$ times meridionally and $q$ times longitudinally. Here a meridian of a solid torus goes around the small cross-section of the doughnut, while the longitudinal direction goes around its central hole. The integers count signed turns, so changing one sign gives the mirror of a link of the torus knot. In this article $p$ is always the meridional count and $q$ the longitudinal count.

An explicit parametric curve in three-dimensional Euclidean space iswhere $R>r>0$. Its cylindrical radius $\rho =\sqrt{x^2+y^2}$ satisfies $(\rho -R{)}^2+z^2=r^2$, the equation of the supporting torus. Its meridional angle is $pt$ and its longitudinal angle is $qt$: this checks the convention directly rather than relying on the picture. Its squared speed isIf two parameter values give the same point, their difference $\delta$ satisfies $p\delta ,q\delta \in 2\pi \mathbb{Z}$. Bézout's identity for the coprime integers $p,q$ then gives $\delta \in 2\pi \mathbb{Z}$. Thus the loop closes and is an injective function on the parameter circle, with nonzero derivative; it is a smooth embedding.

For example, $R=2$, $r=0.7$, $p=2$, $q=3$ produces a trefoil knot. During one circuit the meridional angle changes by $4\pi$ and the longitudinal angle by $6\pi$. The supporting torus is a guide for construction, not a constraint on later ambient isotopies: the knot may move off it. This family makes arithmetic visible as geometry: the coprime integers determine whether a single traversal visits the whole loop without repetition.

The coprime integers hypothesis matters. If the greatest common divisor is $d=\gcd (|p|,|q|)>1$, the same parametric curve retraces a single component $d$ times; it does not automatically draw $d$ different components. The corresponding torus link is instead the full set of points on the torus whose angles satisfy $q\varphi -p\theta \in 2\pi \mathbb{Z}$, with meridional angle $\varphi$ and longitudinal angle $\theta$. It has $d$ disjoint components, each with winding counts $(p/d,q/d)$. For instance, the torus link $T_{2,2}$ consists of two parallel $(1,1)$ loops and is a Hopf link, whereas substituting $(2,2)$ into ${\gamma }_{p,q}$ simply traces one of those loops twice.

The divisor sum of Euler's totient function is the identity arithmetic function:Equivalently, $\phi \ast 1=\mathrm{id}$ under Dirichlet convolution, with $1$ the constant-one arithmetic function. ^[ref]

To see why, group the integers $j=1,\dots ,n$ by $h=\gcd (j,n)$. Each $h$ is a positive integer divisor of $n$. Within its group, $j=ha$ with $1\le a\le n/h$ and $\gcd (a,n/h)=1$, by the scaling identity for greatest common divisors. Dividing by $h$ gives a bijection with the integers counted by $\phi (n/h)$. The groups form a partition of a set, soFor $n=12$, the groups with greatest common divisor $h=1,2,3,4,6,12$ are respectively $\{1,5,7,11\},\{2,10\},\{3,9\},\{4,8\},\{6\},\{12\}$. Their sizes are $4,2,2,2,1,1$, totaling $12$. Möbius inversion reverses this divisor sum to recover Euler's totient function from $g(n)=n$.

For a semiprime $N=pq$ with distinct known primes, this quantum state is uniform over the reduced residue classes. With $n$ binary digits in $N$, its fraction among all $2^n$ computational labels is $a=(p-1)(q-1)/2^n\ge 1/4$. An ancilla qubit whose one probability is $1/(4a)$ dilutes the joint good probability to exactly $1/4$. A reversible range-and-greatest common divisor predicate marks the good labels with ancilla qubit one, and a single exact amplitude amplification iteration produces $|{\xi }_N⟩|1⟩$. Known prime factors provide the cardinality; the membership test itself needs only $N$. Polynomial gate complexity presumes ideal one-qubit rotations at the specified amplitudes.

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]

Consider an original abstract data type whose values are rational numbers. Store an ordered pair $(n,d)$ of integers, interpreted as a numerator and a denominator. Use exact, unbounded integer arithmetic in this model. The storage space, representation invariant and abstraction function areThe representation invariant does not require a positive denominator or a reduced fraction. Consequently $(1,2)$, $(2,4)$ and $(-1,-2)$ are three different concrete states with the same abstract rational number $1/2$. Similarly, $(0,1)$ and $(0,-3)$ both represent zero. The abstraction function is a surjection: every rational number has an integer numerator and a nonzero integer denominator, so every abstract value has a representation.

Both $(1,0)$ and $(0,0)$ violate the representation invariant; neither has a value under this abstraction function. By contrast, $(2,4)$ is completely valid. Simplifying it to the reduced fraction $1/2$ may be a useful implementation choice, but is not a requirement of this representation. Strengthening the representation invariant to require a positive denominator and a greatest common divisor of one would choose a unique stored representative for each rational number; every constructor and operation would then need to establish that stronger condition.

Equality of abstract values is therefore different from equality of stored ordered pairs. For valid $(a,b)$ and $(c,d)$,This defines an equivalence relation on $R_{\mathrm{valid}}$: each equivalence class contains all representations of one rational number. An API equality operation can compare these cross-products using exact integer arithmetic. Comparing the two fields individually would incorrectly declare $(1,2)$ and $(2,4)$ unequal as rational numbers. The general reason is that an abstraction function may map several concrete values to one abstract value.^[ref]