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By the time the candidate $n$ is emitted, every positive integer $m\le n$ has been examined. The straightforward divisor routine makes exactly $m$ remainder tests on $m$. Consequently the total number isand the time is $\Theta (n^2)$ when remainder tests and the other fixed-width operations have constant cost and printing is cheap. Emitting only record candidates does not avoid computing the divisor count of the intervening candidates.

The $n$th pentagonal number is $P_n=n(3n-1)/2$ for a positive integer $n$, giving $1,5,12,22,\dots$. A positive integer $x$ is pentagonal exactly when $24x+1$ is a perfect square and $(1+\sqrt{24x+1})/6$ is an integer, by the defining quadratic equation.

The ratio of the sum of the positive divisors of a positive integer to that integer. For coprime $a,b$, the ratio is multiplicative because the divisor function $\sigma$ is multiplicative. The contribution of a prime power $p^e$ is $(1+p+\cdots +p^e)/p^e$, so a prescribed quotient can be searched through its remaining numerator and denominator.

A perfect number is a positive integer equal to the sum of all its positive proper divisors. The divisor itself is excluded: including it gives the equivalent condition $\sigma (n)=2n$ for the sum-of-divisors function. The smallest examples are $6=1+2+3$ and $28=1+2+4+7+14$.

For comparison, $8$ has proper divisors $1,2,4$, whose sum is $7$: it is a deficient number. The proper divisors of $12$ are $1,2,3,4,6$, whose sum is $16$: it is an abundant number. The proper-divisor sum classifies these examples with the same rule, and gives $s(1)=0$, so one is not a perfect number.

The interesting question is how such balances can be constructed without testing every smaller integer as a possible divisor. The Euclid-Euler theorem answers it for even integers: a Mersenne prime $2^p-1$ gives $2^{p-1}(2^p-1)$, and every even perfect number arises this way. For $p=2,3,5,7$, this produces $6,28,496,8128$. The connection links divisor sums to primality testing for a particular family, while keeping the necessary hypotheses visible.^[ref]

Euclid's Elements already gives the construction in Book IX, Proposition 36: add a succession of powers of two, and if their sum is prime, multiply by the final power of two. The modern finite geometric series $1+2+\cdots +2^{p-1}=2^p-1$ makes the same pattern transparent. This original statement supplies the forward construction, rather than the converse classification of all even perfect numbers.^[ref]

A point $x$ is periodic for an interval map $F$ when $F^n(x)=x$ for some positive integer $n$. Its least such $n$ is its period, and the finite setis its periodic orbit.

For an interval map, this is the usual periodic point condition. Its least positive return time is the period.

Place $N$ objects into $m\ge 1$ boxes, assigning each object to exactly one box. Here $N$ is a nonnegative integer and $m$ is a positive integer. The pigeonhole principle says that if $N>m$, some box contains at least two objects. Equivalently, a function from an $N$-element finite set to an $m$-element finite set cannot be an injective function when $N>m$. The boxes may stand for categories rather than physical containers. ^[ref]

For thirteen objects and twelve boxes, a box containing at least two is unavoidable. Thinking of thirteen people and their birth months gives the same conclusion: at least two share a birth month. Twelve people could occupy the twelve months one apiece, so thirteen is the smallest number that forces a shared month. The generalized pigeonhole principle quantifies how many objects must share a box when there are still more objects.

The useful surprise is that the pigeonhole principle proves existence without needing to locate the objects involved. Choosing a good assignment rule is often the creative step: equal labels can mean equal remainders, nearby points, or repeated states of an algorithm. Once the rule is chosen, a short count can force a relationship among a much larger collection of objects. The pigeonhole principle itself does not name the crowded box. When the objects and their assignments are explicitly available, a simple algorithm finds one: scan the objects, keep the first object seen in each box, and stop when a later object enters an occupied box. At most $m+1$ scanned objects are needed when $N>m$.

A geometric example. Any five points in a square of side one contain two whose Euclidean distance is at most $1/\sqrt{2}$. Give each point Cartesian coordinates $(x,y)\in [0,1{]}^2$. Divide both ranges into the intervals $[0,1/2)$ and $[1/2,1]$, making four cells. Thus $x<1/2$ belongs to the left half and $x\ge 1/2$ to the right half; $y<1/2$ belongs to the lower half and $y\ge 1/2$ to the upper half. This convention assigns every boundary point to exactly one cell.

The pigeonhole principle puts two of the five points in the same cell. Their differences in Cartesian coordinates have magnitude at most $1/2$, so the Pythagorean theorem gives their Euclidean distance $d$ the boundFor the plotted points $(0.1,0.1)$, $(0.4,0.4)$, $(0.8,0.2)$, $(0.2,0.8)$ and $(0.8,0.8)$, the first two occupy the lower-left cell and have Euclidean distance $\sqrt{0.3^2+0.3^2}=\sqrt{0.18}\approx 0.424264$. These Cartesian coordinates exhibit the forced pair; the argument applies to every placement of five points.

An arithmetic example. Assign eight integers to their seven residue classes modulo the integer modulus $7$. The pigeonhole principle gives two with the same residue class, so $7$ is an integer divisor of their difference. In $\{2,5,9,12,16,20,24,27\}$, for example, $2$ and $9$ satisfy the integer congruence $9\equiv 2(\mathrm{mod}7)$, and $9-2=7$. Likewise, assigning fractional parts to short intervals gives the Dirichlet approximation theorem; its existing article explains the approximation bound.

In this positive integer context, a power of two is $2^k$ for a nonnegative integer $k$, giving $1,2,4,8,\dots$. Allowing negative or real exponents gives a broader meaning of exponentiation, which is not intended in the integer divisibility criteria here.

A positive integer $n$ such that every positive integer at most $n$ is a sum of distinct divisors of $n$. For example, $6$ is practical because its divisors $1,2,3,6$ represent every value from $1$ through $6$.

Primality is the property of a positive integer being a prime number. Primality testing determines this property; integer factorization asks for factors and is a different computational task.

A prime number is a positive integer $p>1$ whose only positive divisors are $1$ and $p$. The first ten prime numbers are $2,3,5,7,11,13,17,19,23,29$. In contrast, $12=3\cdot 4$ is a composite number because it has a positive divisor strictly between one and itself. The integer one is neither a prime number nor a composite number: the lower bound $p>1$ is part of the definition. Equivalently, the divisor function satisfies $\tau (p)=2$ for a prime number, whereas $\tau (1)=1$ and $\tau (n)\ge 3$ for every composite number $n$.

Prime numbers are the building blocks of positive integers under multiplication. The fundamental theorem of arithmetic says that every integer greater than one is a product of prime numbers, with a unique collection of factors when their order is ignored. For example, $12=2\cdot 2\cdot 3$ and $18=2\cdot 3\cdot 3$. Their common factors reveal the greatest common divisor $6$, and the combined maximum powers reveal the least common multiple $36$. A complicated divisibility question becomes a comparison of prime exponents.

There is also a useful finite test. If a composite number $n$ equals $ab$ with $1<a\le b$, then $a^2\le n$. Therefore some prime factor is at most $\sqrt{n}$, and trial division only needs to check prime numbers up to that square root. For $97$, checking $2,3,5,7$ suffices; none divides $97$, so it is a prime number. For $91$, the same test finds $7$, giving $91=7\cdot 13$. To find many prime numbers together, the sieve of Eratosthenes reuses the work by marking multiples.

The division with remainder calculations for $97$ make the test explicit:Since $9^2<97<10^2$, these are all the prime numbers at most its square root, and every remainder is nonzero. Thus the calculation certifies a prime number, without testing the integers $8,9,10,\dots ,96$ individually. In contrast, $91=7\cdot 13+0$ gives a prime factor and proves that $91$ is a composite number.

The definition has a long history. Book VII of Euclid's Elements distinguishes prime numbers from composite numbers through their possible measures; Book IX, Proposition 20 establishes that no finite list exhausts the prime numbers. The modern argument under Euclid's theorem needs only divisibility, so it proves an infinite conclusion without estimating how frequently prime numbers occur. ^{[ref] [ref]}.

The prime omega function $\omega (n)$ counts the distinct prime factors of a positive integer $n$, so it is the cardinality of the prime support of an integer $P(n)$. For a prime factorization with distinct prime numbers $p_1,\dots ,p_r$ and positive exponents,At $n=1$ there are no prime factors, so the empty sum gives $\omega (1)=0$. The total number of prime factors instead gives $\Omega (n)={\sum }_i{a}_i$. ^[ref]

For $72=2^3\cdot 3^2$, the prime support of an integer is $\{2,3\}$ and $\omega (72)=2$. The comparison examples give $\omega (30)=3$, $\omega (32)=1$ and $\omega (36)=2$. Raising an existing prime factor to a higher power does not affect the prime omega function: $\omega (p^k)=1$ for a prime number $p$ and $k\ge 1$. Thus the prime omega function is a strongly additive arithmetic function, withThe reason is that coprime integers have disjoint prime supports of an integer. For arbitrary positive integers, their prime support of an integer combines by set union, and the shared prime factors are exactly those of the greatest common divisor. Counting this set union gives the more general identityFor $a=12$, $b=18$, both prime supports of an integer are $\{2,3\}$ and $\gcd (a,b)=6$, so $\omega (216)=2+2-2=2$. In particular $\omega (4)=1\ne 2=\omega (2)+\omega (2)$: the prime omega function is not a completely additive arithmetic function.

The prime omega function is useful for counting choices involving distinct prime factors. Every squarefree divisor of $n$ chooses a subset of its prime support of an integer, so there are $2^{\omega (n)}$ squarefree divisors. For $72$, this gives $2^2=4$, namely $1,2,3,6$. Also $\omega (\mathrm{rad}(n))=\omega (n)$, because the radical of an integer keeps exactly that prime support of an integer. The total number of prime factors records what this count discards, and the two counts agree exactly on squarefree integers.

A prime sum is a summation indexed by prime numbers, for example ${\sum }_{p\le x}1/p$. The indexing restriction distinguishes it from a sum over all positive integers.

For a prime number $p$ and a positive integer $n$, the prime valuation is the exponent of $p$ in the prime factorization of $n$, equivalentlyThus $v_2(360)=3$ because $8$ divides $360$ but $16$ does not; $v_3(360)=2$, $v_5(360)=1$ and $v_7(360)=0$. In particular $v_p(1)=0$. Multiplication adds prime valuations, $v_p(ab)=v_p(a)+v_p(b)$, and the fundamental theorem of arithmetic makes comparison of prime valuations an exact test for divisibility between positive integers. This definition extends to a nonzero integer by taking its absolute value; for zero one commonly sets $v_p(0)=+\infty$.

A Pythagorean triple $(a,b,c)$ whose coordinates have greatest common divisor 1. Its two legs have opposite integer parity, and every Pythagorean triple is a unique positive integer multiple of a primitive Pythagorean triple up to the ordering of its legs.

A positive integer $N$ that is both the sum and the product of one multiset of $k$ positive integers. For factors at least $2$ with product $P$, sum $S$, and count $t$, appending $P-S$ ones gives $k=t+P-S$ without changing the product. For $k\ge 2$, the example $2,k$ with $k-2$ ones proves that the smallest product-sum value is at most $2k$.

A positive integer $n$ for which some division $n=dq+r$ with integers $d,q,r>0$ and $r<d$ has $d,q,r$ as consecutive terms of a geometric progression in some order. Writing a reduced common ratio as $a/b$ with $a>b>0$ and $\gcd (a,b)=1$, the three terms have the form $t{b}^2,tab,t{a}^2$ for a positive integer $t$.

Here a proper divisor of a positive integer $n$ means a positive integer divisor other than $n$ itself. Thus $1$ has no proper divisors, and the proper divisors of $12$ are $1,2,3,4,6$.

The equation $a^2+b^2=c^2$. Its positive integer solutions are Pythagorean triples; primitive Pythagorean triples are those with no common factor greater than 1.

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 $36\mid 72$.

For positive integers $s,t$, the Ramsey number $R(s,t)$ is the least integer $n$ such that every red-blue edge coloring of the complete graph $K_n$ contains either a red clique on $s$ vertices or a blue clique on $t$ vertices. The complete graph $K_n$ has all $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ unordered pairs as edges; every edge gets exactly one color, with no requirement that incident edges have different colors. The definition and its graph interpretation appear in D. Yogeshwaran's Discrete Mathematics notes, section 5.3.

For example, $R(3,3)$ asks how many vertices force a monochromatic triangle in a graph. Swapping colors gives $R(s,t)=R(t,s)$. Also $R(2,t)=t$: either a red edge exists, or all edges are blue. On $t-1$ vertices, an all-blue coloring avoids both a red edge and a blue $t$-clique. Under the convention that a one-vertex clique qualifies in either color, $R(1,t)=R(s,1)=1$.

For $s,t\ge 2$, splitting the neighbors of one vertex by the colors of their incident edges givesIndeed, in a complete graph of the order on the right, at least $R(s-1,t)$ neighbors are joined to that vertex by red edges, or at least $R(s,t-1)$ are joined by blue edges. In the red case, the defining property of the smaller Ramsey number supplies either a blue $t$-clique among the neighbors or a red $(s-1)$-clique that extends by the chosen vertex; the blue case is symmetric. Starting from the one-vertex base cases, this recursively proves finite existence. The binomial upper bound for a Ramsey number packages the resulting estimate. ^[ref] present the neighbor argument and the sharp Ramsey number for triangles.

Equivalently, every graph on at least $R(s,t)$ vertices has an $s$-clique or a $t$-element independent set: color its edges red and the edges of its complement graph blue. This is the Ramsey theorem in graph theory. Larger monochromatic patterns have their own thresholds, such as a graph Ramsey number, matching Ramsey number or triangle-packing Ramsey number; these ask different questions from finding one clique.