Using Bézout's identity

In modern mathematics, a common proof involves Bézout's identity, which was unknown at Euclid's time.^[12] Bézout's identity states that if x and y are coprime integers (i.e. they share no common divisors other than 1 and −1) there exist integers r and s such that

${\displaystyle rx+sy=1.}$

Let a and n be coprime, and assume that n|ab. By Bézout's identity, there are r and s such that

${\displaystyle rn+sa=1.}$

Multiply both sides by b:

${\displaystyle rnb+sab=b.}$

The first term on the left is divisible by n, and the second term is divisible by ab, which by hypothesis is divisible by n. Therefore their sum, b, is also divisible by n.

By induction

The following proof is inspired by Euclid's version of Euclidean algorithm, which proceeds by using only subtractions.

Suppose that ${\displaystyle n\mid ab}$ and that n and a are coprime (that is, their greatest common divisor is 1). One has to prove that n divides b. Since ${\displaystyle n\mid ab,}$ there is an integer q such that ${\displaystyle nq=ab.}$ Without loss of generality, one can suppose that n, q, a, and b are positive, since the divisibility relation is independent from the signs of the involved integers.

For proving this by strong induction, we suppose that the result has been proved for all positive lower values of ab.

There are three cases:

If n = a, coprimality implies n = 1, and n divides b trivially.

If n < a, one has

${\displaystyle n(q-b)=(a-n)b.}$

The positive integers a – n and n are coprime: their greatest common divisor d must divide their sum, and thus divides both n and a. It results that d = 1, by the coprimality hypothesis. So, the conclusion follows from the induction hypothesis, since 0 < (a – n) b < ab.

Similarly, if n > a one has

${\displaystyle (n-a)q=a(b-q),}$

and the same argument shows that n – a and a are coprime. Therefore, one has 0 < a (b − q) < ab, and the induction hypothesis implies that n − a divides b − q; that is, ${\displaystyle b-q=r(n-a)}$ for some integer. So, ${\displaystyle (n-a)q=ar(n-a),}$ and, by dividing by n − a, one has ${\displaystyle q=ar.}$ Therefore, ${\displaystyle ab=nq=anr,}$ and by dividing by a, one gets ${\displaystyle b=nr,}$ the desired result.

Proof of Elements

Euclid's lemma is proved at the Proposition 30 in Book VII of Euclid's Elements. The original proof is difficult to understand as is, so we quote the commentary from Euclid (1956, pp. 319–332).

Proposition 19

If four numbers be proportional, the number produced from the first and fourth is equal to the number produced from the second and third; and, if the number produced from the first and fourth be equal to that produced from the second and third, the four numbers are proportional.^{[note 3]}

Proposition 20

The least numbers of those that have the same ratio with them measures those that have the same ratio the same number of times—the greater the greater and the less the less.^{[note 4]}

Proposition 21

Numbers prime to one another are the least of those that have the same ratio with them.^{[note 5]}

Proposition 29

Any prime number is prime to any number it does not measure.^{[note 6]}

Proposition 30

If two numbers, by multiplying one another, make the same number, and any prime number measures the product, it also measures one of the original numbers.^{[note 7]}

Proof of 30

If c, a prime number, measure ab, c measures either a or b.
Suppose c does not measure a.
Therefore c, a are prime to one another. ［VII. 29］
Suppose ab＝mc.
Therefore c : a ＝ b : m. ［VII. 19］
Hence ［VII. 20, 21］ b＝nc, where n is some integer.
Therefore c measures b.
Similarly, if c does not measure b, c measures a.
Therefore c measures one or other of the two numbers a, b.
Q.E.D.^[18]