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If, for every $\eta >0$, the smooth numbers up to $T$ with all prime factors at most $T^{\eta }$ have density at least $\kappa (\eta )>0$, then for fixed $\epsilon >0$ there are only $O_{\epsilon }(1)$ primes $p\le N$ with $n(p)>N^{\epsilon }$. Apply the variance form of the large sieve to the indicator function of the $N^{\epsilon }$-smooth numbers up to $N^2$. For every exceptional prime, all these numbers are nonzero quadratic residues, so half its nonzero residue classes are empty. Their variance contribution is at least one third of the square of the total count. The large sieve bounds the sum of contributions by a constant times $N^2$ times that count. The density assumption with $\eta =\epsilon /2$ finishes the argument.

A composite number is an integer greater than one that is not a prime number. Equivalently, it has a factorization $n=ab$ with integers $a,b>1$. Thus $4=2\cdot 2$, $9=3\cdot 3$, and $12=3\cdot 4$ are composite numbers; one is neither a composite number nor a prime number.

Every composite number has a prime factor at most its square root. In $91=7\cdot 13$, seven is the smaller factor and lies below $\sqrt{91}$; in $49=7^2$, equality occurs. This square-root bound makes trial division finite and lets the sieve of Eratosthenes stop marking with prime numbers beyond $\sqrt{N}$ when listing prime numbers up to $N$.

A congruence of squares says that two integers $a,b$ have the same square modulo a positive integer $N$. Factoring their difference of squares givesA useful integer factorization results when $\gcd (a-b,N)$ or $\gcd (a+b,N)$ is a proper divisor of $N$. Divisibility of a product does not generally force divisibility of either factor by all of $N$: different prime factors of $N$ can divide different parts of that product.

For $N=77$, $20^2-13^2=231=3\cdot 77$. ThereforeThus $77=7\cdot 11$. The prime seven divides the difference, while eleven divides the sum. These computations already give a complete prime factorization because seven and eleven are prime numbers.

The Chinese remainder theorem explains the hidden sign choices. The four congruence classes solving $z^2\equiv 15(\mathrm{mod}77)$ have representatives $13,20,57,64$. Modulo seven each is $1$ or $-1$; modulo eleven each is $2$ or $-2$. Two independent sign choices produce four roots. The inputs thirteen and twenty agree modulo seven and disagree by sign modulo eleven, allowing their sum and difference to separate the two prime factors.

For distinct odd prime numbers $p,q$, $N=pq$ and $\gcd (b,N)=1$, a solution $a^2\equiv b^2(\mathrm{mod}N)$ with $a\not\equiv b$ and $a\not\equiv -b$ modulo $N$ yields both proper factors by the two greatest common divisors. For a general composite integer, a proper divisor need not itself be prime; further factoring can remain. If $a\equiv b$ or $a\equiv -b(\mathrm{mod}N)$ in the coprime case, the two calculations return $N$ and one instead. A factor base helps generate relations, but does not ensure every resulting relation is useful.

For historical context, Pomerance describes Kraitchik's 1920s enhancement of the difference-of-squares approach associated with Fermat: seek a difference that is a multiple of $N$, rather than exactly $N$. His account also notes antecedents in Gauss and Seelhoff, so this is not an attribution of the whole idea to a single inventor. ^[ref].

Two integers $a,b$ are coprime when their greatest common divisor is one:For positive integers, this says that the only positive common divisor is one. For signed integers, use $\gcd (a,b)=\gcd (|a|,|b|)$, with absolute values; changing either sign leaves coprimality unchanged. With the convention $\gcd (0,0)=0$, zero is coprime precisely to $1,-1$, because $\gcd (a,0)=|a|$. Both $1,-1$ are coprime to every integer. ^[ref]

The word concerns shared prime factors, not whether either input is a prime number. For instance $8=2^3$ and $9=3^2$ are coprime composite numbers, even though neither is a square-free integer. Conversely, $6=2\cdot 3$ and $10=2\cdot 5$ are square-free integers but are not coprime. Consecutive integers are always coprime: any common divisor of $n,n+1$ divides their difference one. Thus $14,15$ provide another pair of coprime composite numbers.

For positive integers, let $P(n)$ be the prime support of an integer. The equivalent test isThe set intersection of the two prime supports of an integer must be the empty set. Indeed, any positive common divisor greater than one has a prime factor, and any shared prime factor is a common divisor greater than one. Hence coprimality depends on which prime factors occur, rather than on their prime valuations; replacing either input by its radical of an integer preserves coprimality.

For a list of at least two integers, distinguish pairwise coprimality, requiring every pair to be coprime, from joint coprimality, requiring only the greatest common divisor of the entire list to be one. For example,No prime factor belongs to all three prime supports of an integer, yet each pair shares one. By contrast, $4,9,25$ have pairwise coprimality. The distinction is presented by Poritz under the names pairwise and mutually relatively prime. ^[ref]

One use is simplifying fractions. For a fraction $a/b$ with integer entries and $b>0$, divide its numerator and denominator by $d=\gcd (a,b)$ to obtain a reduced fraction with coprime entries. For example, $42/30=7/5$ since $d=6$, and $-42/30=-7/5$ uses the same $d$. A further common divisor would make $d$ too small, so the reduction is complete. In particular, the reduced fraction for zero is $0/1$.

Another use is reversible modular arithmetic: for an integer modulus $m>1$, an integer $a$ has a modular inverse precisely when $a,m$ are coprime. The proof outline uses Bézout's identity: $ua+vm=1$ makes $u$ a modular inverse, and any inverse relation $ua-km=1$ forces every common divisor to divide one. For example, $3\cdot 5-2\cdot 7=1$ gives $5^{-1}\equiv 3(\mathrm{mod}7)$, whereas $6$ has no modular inverse modulo $10$. Thus coprimality is the exact condition allowing division by a residue in modular arithmetic. ^[ref]

For a reduced fraction $p/q$ with $0<p<q$, $q>1$, and $q,10$ coprime integers, its decimal representation repeats immediately after the decimal point. The shortest repeating block has lengthHere $\mathrm{ord}$ denotes multiplicative order, and $\phi$ denotes Euler's totient function. Changing the numerator preserves the decimal period as long as the fraction remains reduced.^[ref]

For example, long division of $1/7$ gives the remainders $1,3,2,6,4,5,1$, with emitted digits $1,4,2,8,5,7$. ConsequentlyEach remainder determines both the next digit and the next remainder; returning to one restarts the same six steps.

More generally, initialize long division with $r_0=p$. At each step, division with remainder givesso $r_j\equiv p10^j(\mathrm{mod}q)$. Since $p,q$ are coprime integers, the modular inverse of $p$ shows that $r_L=r_0$ exactly when $10^L\equiv 1(\mathrm{mod}q)$. To see why the digit block cannot repeat sooner than the remainders, a purely repeating decimal with a block of $L$ digits represents $D/(10^L-1)$ for an integer $D$. A reduced fraction $p/q$ with that value forces $q\mid 10^L-1$. Hence the shortest digit repeat and the multiplicative order coincide.

Knowing the decimal period predicts how many digits must be computed and stored: one block determines the whole repeating decimal. It also constructs the block directly. With $L={\mathrm{ord}}_q(10)$, the integeris the block, padded on the left with zeros to $L$ digits. For $1/11$, $L=2$ and $D=9$, so the block is $09$ and $1/11=0.\overline{09}$. Reduction matters: $7/21=1/3=0.\overline{3}$ has decimal period one, whereas ${\mathrm{ord}}_{21}(10)=6$.

If the reduced denominator contains prime factors two or five, ten has no multiplicative order modulo that whole denominator. WriteThe decimal preperiod has exactly $k_0$ digits. If $m>1$, the remaining repeating decimal has decimal period ${\mathrm{ord}}_m(10)$. For instance, $28=2^2\cdot 7$ gives $1/28=0.03\overline{571428}$: two prefix digits, then six repeating digits. If $m=1$, the decimal representation is a terminating decimal with exactly $k_0$ fractional places. We use the representation ending in zeros, rather than its alternative ending in nines; for example $1/40=0.025$ terminates after three places.

A short derivation explains both cases. After $k$ digits, multiplying the reduced fraction by $10^k$ and discarding its integer part leaves a reduced fraction whose denominator is $q/\gcd (q,10^k)$. This denominator first becomes coprime to ten at $k=k_0$. It then equals $m$, giving an immediate repeat when $m>1$, or an integer when $m=1$. The same long division argument works for positional notation in any base $b\ge 2$: for a reduced denominator $q>1$ coprime to $b$, the shortest repeating block has length ${\mathrm{ord}}_q(b)$.

The Dirichlet convolution of arithmetic functions $f,g$ is the arithmetic functionIn the first sum, $d$ runs over the positive integer divisors of $n$; the second uses ordered pairs of positive integers whose product is $n$. Both sums are finite. Each pair contributes one weight from $f$ and one from $g$, so Dirichlet convolution combines information attached to complementary factors. It turns a divisor sum into an operation on arithmetic functions. ^[ref]

For $n=12$ the six ordered pairs are $(1,12),(2,6),(3,4),(4,3),(6,2),(12,1)$. The constant-one arithmetic function gives one per pair, while the identity arithmetic function weights the first factor by its size:Here $\tau$ is the divisor function and ${\sigma }_1$ is the sum-of-divisors function. By comparison, the pointwise product of arithmetic functions gives $(1\cdot 1)(12)=1$ and $(\mathrm{id}\cdot 1)(12)=12$.

Commutativity follows by swapping the two entries of every ordered pair. Associativity follows by reindexing either grouping over the same ordered tuples of three positive integers:For example, $((1\ast 1)\ast 1)(12)$ counts ordered tuples of three positive integers with product $12$. Distribute the two occurrences of the prime factor $2$ among three entries in $\left(\genfrac{}{}{0ex}{}{2+2}{2}\right)=6$ ways, and the one occurrence of the prime factor $3$ in $\left(\genfrac{}{}{0ex}{}{1+2}{2}\right)=3$ ways. These binomial coefficients are the stars and bars counts of exponent choices in the nonnegative integers; prime factorization therefore gives $6\cdot 3=18$ such tuples. Directly, the same Dirichlet convolution is ${\sum }_{d\mid 12}\tau (d)=1+2+2+3+4+6=18$.

The Identity for Dirichlet convolution is $\epsilon$, and the Möbius function reverses a divisor sum because $\mu \ast 1=\epsilon$. This makes Dirichlet convolution useful both for counting factorizations and for recovering an arithmetic function from its divisor sum.

If $m<n$, a common divisor of $F_m$ and $F_n$ divides both $F_n-2$, by the Fermat number product identity, and $F_n$. It therefore divides two. Both Fermat numbers are odd integers, so their greatest common divisor is one. Each term's prime factors are consequently different from the prime factors of every other term.

A divisor of an integer $n$ is an integer $d$ satisfying the divisibility relation $d\mid n$. Equivalently, $n=dk$ for some integer $k$. The word positive matters when listing or counting divisors: unless signs are restricted, both $d$ and $-d$ qualify.

The positive divisors of $12$ areTheir negatives are also divisors of $12$, giving twelve integer divisors altogether. The same list of positive divisors works for $-12$. One has only the positive divisor $1$, while zero has every integer as a divisor, including zero under our divisibility convention. A nonzero integer never has zero as a divisor.

For a positive integer $n$, the positive divisors come in pairs $(d,n/d)$ whose product is $n$. For $12$, the pairs are $(1,12)$, $(2,6)$ and $(3,4)$. Each pair contains a divisor at most $\sqrt{n}$, so searching through candidate divisors up to the square root suffices to find all pairs. A pair repeats its own member only when $n$ is a perfect square; for $36$, the middle pair is $(6,6)$. Consequently a positive integer has an odd number of positive divisors exactly when it is a perfect square.

The prime factorization gives another complete description. Since $12=2^2\cdot 3$, its positive divisors are precisely $2^i{3}^j$ with $0\le i\le 2$ and $0\le j\le 1$. These independent choices give $(2+1)(1+1)=6$ positive divisors, the value of the divisor function $\tau (12)$. A prime factor is one particular kind of divisor; $4$, $6$ and $12$ show that divisors need not be prime numbers.

Thinking in pairs connects exact grouping with integer factorization: twelve objects can form three groups of four or four groups of three. Thinking in shared divisors gives the greatest common divisor; for $12$ and $18$, the positive common divisors are $1,2,3,6$, and the largest is $6$.

The divisor function $\tau (n)={\sum }_{d\mid n}1$ counts the positive integer divisors of a positive integer $n$. In particular $\tau (1)=1$, because $1$ has just one positive integer divisor. More generally the notation ${\sigma }_k(n)={\sum }_{d\mid n}{d}^k$ gives $\tau ={\sigma }_0$ and the sum-of-divisors function ${\sigma }_1$. Thus the divisor function counts the positive integer divisors, while ${\sigma }_1$ adds their sizes. ^[ref]

If the prime factorization is $n={\prod }_{k=1}^r{p}_k^{a_k}$ with distinct prime numbers $p_k$ and positive exponents $a_k$, thenThe short proof is a count: each positive integer divisor has a unique form ${\prod }_{k=1}^r{p}_k^{b_k}$, where $0\le b_k\le a_k$. There are $a_k+1$ choices at each prime number, and choosing them gives every positive integer divisor exactly once. For $n=1$ there are no prime factors and the product is $1$.

For $12=2^2\cdot 3$, choose $i\in \{0,1,2\}$ and $j\in \{0,1\}$ to obtain $2^i{3}^j$. The six values are $1,2,4,3,6,12$; sorted, the positive integer divisors are $1,2,3,4,6,12$. Therefore $\tau (12)=(2+1)(1+1)=6$. The diagram makes the independent exponent choices visible.

This count is useful because a known prime factorization gives $\tau (n)$ without testing every smaller positive integer for divisibility. It also proves the divisor function is a multiplicative function: coprime integers have disjoint prime factors, so their exponent choices combine. For example $\tau (6)=4=\tau (2)\tau (3)$, but $\tau (4)=3\ne 4=\tau (2{)}^2$, so it is not a completely multiplicative function. Equivalently, $\tau =1\ast 1$ under Dirichlet convolution, counting ordered pairs of positive integers whose product is $n$.

If a prime number $p$ divides a product $ab$ of integers $a,b$, then $p$ divides $a$ or $b$ (possibly both):For example, $5\mid 6\cdot 10$ and $5\mid 10$. The hypothesis that $p$ is a prime number matters: $6\mid 2\cdot 3$, although six divides neither factor. When $p$ does not divide $a$, the greatest common divisor of $p$ and $a$ is one; multiply Bézout's identity $up+va=1$ by $b$ to see that $p\mid b$. Repeated application makes this the uniqueness step in the fundamental theorem of arithmetic. ^[ref].

For a completely computed example, $7\mid 20\cdot 21$ because $20\cdot 21=420=7\cdot 60$, while $20=7\cdot 2+6$ shows that seven does not divide twenty. Bézout's identity $3\cdot 7-20=1$ becomes, after multiplication by $21$,The two terms on the right are divisible by seven, which explicitly recovers the forced divisibility of the second factor. Both alternatives can hold: $7\mid 14\cdot 21$, and seven divides both $14$ and $21$.

In fact the property characterizes prime numbers among the integers greater than one. If $c$ is a composite number, choose $c=ab$ with $1<a,b<c$; then $c\mid ab$, but $c$ divides neither $a$ nor $b$. For several factors, repeated application of Euclid's lemma says that a prime number dividing their product divides at least one factor. In a product of prime numbers, that factor must equal it, because a prime number has no other prime factor. This is what prevents a different collection of prime factors from representing the same positive integer in the fundamental theorem of arithmetic.

Book VII, Proposition 30 of Euclid's Elements states the result in terms of a prime number measuring a product; its proof uses coprimality and ratios. The calculation using Bézout's identity above is a modern way to explain the same divisibility statement. ^[ref]