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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]

Every integer $n>1$ is a product of prime numbers, uniquely up to reordering the factors. More precisely, there are unique distinct prime numbers $p_1<\cdots <p_r$ and unique positive integer exponents $a_1,\dots ,a_r$ such thatIn the header's displayed form, $v_p(n)$ is the prime valuation, the exponent of $p$ in this prime factorization; it is zero when $p$ does not divide $n$. Reordering or regrouping the prime factors does not give a different prime factorization.

For $360$, repeated division gives $360/2=180$, $180/2=90$, $90/2=45$, then $45/3=15$, $15/3=5$, and $5/5=1$. ThusIts nonzero prime valuations are $v_2(360)=3$, $v_3(360)=2$ and $v_5(360)=1$. Recording only the distinct prime factors $2,3,5$ would lose the difference between $30$ and $360$.

This turns divisibility into coordinatewise comparison of nonnegative exponents. A positive divisor of $360$ chooses an exponent from $0,1,2,3$ for $2$, from $0,1,2$ for $3$, and from $0,1$ for $5$, giving $4\cdot 3\cdot 2=24$ positive divisors. The greatest common divisor takes the smaller exponent at each prime number; the least common multiple takes the larger. The theorem is a structural description, not an assertion that finding the factors of every large integer is computationally easy.

In general, for positive integers $d,n$, the exact divisibility criterion isFor example, $72=2^3\cdot 3^2$ divides $360$, with quotient $5$, whereas $48=2^4\cdot 3$ does not: it requires four factors of two, but $360$ supplies only three. Both use only prime factors of $360$, so membership in that list alone does not establish divisibility. Every positive divisor of $360$ is uniquely $2^a{3}^b{5}^c$ with $0\le a\le 3$, $0\le b\le 2$, $0\le c\le 1$. Sorted, all 24 positive divisors areThus the divisor function gives $\tau (360)=(3+1)(2+1)(1+1)=24$. These include divisors that are composite numbers, as well as the prime factors and one.

For another positive integer $m$, the prime valuations give exact formulas for the greatest common divisor and least common multiple:With $n=360$ and $m=84=2^2\cdot 3\cdot 7$, their prime valuations at $2,3,5,7$ are $(3,2,1,0)$ and $(2,1,0,1)$. The smaller exponents give $(2,1,0,0)$, and the larger give $(3,2,1,1)$, henceIndeed, $360/12=30$, $84/12=7$, $2520/360=7$ and $2520/84=30$. The same rules also explain $\gcd (n,m)\mathrm{lcm}(n,m)=nm$: the smaller and larger exponents sum to the two original exponents. Here $12\cdot 2520=360\cdot 84=30240$.

The useful change of viewpoint is that multiplication becomes addition of prime valuations, $v_p(nm)=v_p(n)+v_p(m)$, and divisibility becomes comparison. Many different-looking calculations therefore share one representation. The prime valuations are coordinates for multiplication, rather than digits in positional notation: they do not turn ordinary addition of integers into coordinatewise addition.

Different intermediate splits lead to the same prime ingredients. Starting with $360=36\cdot 10$ gives $36=6\cdot 6$ and $10=2\cdot 5$; starting with $360=8\cdot 45$ gives $8=2\cdot 4$ and $45=5\cdot 9$. Continuing to split the composite numbers produces the same multiset of prime factors: three twos, two threes, and one five. The factorization tree records choices in the calculation, while the prime factorization records the invariant result.

The proof outline explains the uniqueness. Existence follows by splitting a composite number into smaller factors until only prime numbers remain. For uniqueness, Euclid's lemma says that a prime factor of one product must divide one factor of any other prime product equal to it. That factor must be the same prime number; cancel it and repeat. The crucial divisibility step appears in Book VII, Proposition 30 of Euclid's Elements. ^[ref]

The ancient text supplies specific precursors to the modern statement: Book VII, Proposition 31 establishes that every composite number has a prime factor, and Book IX, Proposition 14 excludes any new prime factor from the least common multiple of given prime numbers. These are the propositions being cited, rather than the complete modern exponent formulation of the fundamental theorem of arithmetic. ^{[ref] [ref]}

Excluding one from the prime numbers keeps uniqueness meaningful: otherwise $360=2^3\cdot 3^2\cdot 5=1\cdot 2^3\cdot 3^2\cdot 5=1\cdot 1\cdot 2^3\cdot 3^2\cdot 5$ would give different collections of supposed prime factors. For $n=1$, the natural convention is the empty product, equal to one, with no prime factors and all prime valuations zero.

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.