# Euclid's theorem

Euclid's theorem is a fundamental statement in number theory that asserts that there are infinitely many prime numbers. There are several proofs of the theorem.

## Euclid's proof

Euclid offered a proof published in his work Elements (Book IX, Proposition 20),[1] which is paraphrased here.[2]

Consider any finite list of prime numbers p1p2, ..., pn. It will be shown that at least one additional prime number not in this list exists. Let P be the product of all the prime numbers in the list: P = p1p2...pn. Let q = P + 1. Then q is either prime or not:

• If q is prime, then there is at least one more prime that is not in the list.
• If q is not prime, then some prime factor p divides q. If this factor p were in our list, then it would divide P (since P is the product of every number in the list); but p divides P + 1 = q. If p divides P and q, then p would have to divide the difference[3] of the two numbers, which is (P + 1) − P or just 1. Since no prime number divides 1, p cannot be on the list. This means that at least one more prime number exists beyond those in the list.

This proves that for every finite list of prime numbers there is a prime number not in the list, and therefore there must be infinitely many prime numbers.

Euclid is often erroneously reported to have proved this result by contradiction, beginning with the assumption that the finite set initially considered contains all prime numbers, or that it contains precisely the n smallest primes, rather than any arbitrary finite set of primes.[4] Instead of a proof by contradiction, Euclid's proof shows that every finite list has a particular property. A contradiction is not inferred, but none of the items in the list can have the property of being a divisor of 1.

Several variations on Euclid's proof exist, including the following:

The factorial n! of a positive integer n is divisible by every integer from 2 to n, as it is the product of all of them. Hence, n! + 1 is not divisible by any of the integers from 2 to n, inclusive (it gives a remainder of 1 when divided by each). Hence n! + 1 is either prime or divisible by a prime larger than n. In either case, for every positive integer n, there is at least one prime bigger than n. The conclusion is that the number of primes is infinite.[5]

## Euler's proof

Another proof, by the Swiss mathematician Leonhard Euler, relies on the fundamental theorem of arithmetic: that every integer has a unique prime factorization. If P is the set of all prime numbers, Euler wrote that:

${\displaystyle \prod _{p\in P}{\frac {1}{1-{\frac {1}{p}}}}=\prod _{p\in P}\sum _{k\geq 0}{\frac {1}{p^{k}}}=\sum _{n}{\frac {1}{n}}.}$

The first equality is given by the formula for a geometric series in each term of the product. The second equality is a special case of the Euler product formula for the Riemann zeta function. To show this, distribute the product over the sum:

{\displaystyle {\begin{aligned}\prod _{p\in P}\sum _{k\geq 0}{\frac {1}{p^{k}}}&=\sum _{k\geq 0}{\frac {1}{2^{k}}}\times \sum _{k\geq 0}{\frac {1}{3^{k}}}\times \sum _{k\geq 0}{\frac {1}{5^{k}}}\times \sum _{k\geq 0}{\frac {1}{7^{k}}}\times \cdots \\[8pt]&=\sum _{k,\ell ,m,n,\cdots \geq 0}{\frac {1}{2^{k}3^{\ell }5^{m}7^{n}\cdots }}=\sum _{n}{\frac {1}{n}}\end{aligned}}}

in the result, every product of primes appears exactly once and so by the fundamental theorem of arithmetic the sum is equal to the sum over all integers.

The sum on the right is the harmonic series, which diverges. Thus the product on the left must also diverge. Since each term of the product is finite, the number of terms must be infinite; therefore, there is an infinite number of primes.

## Erdős's proof

Paul Erdős gave a third proof that also relies on the fundamental theorem of arithmetic. First note that every integer n can be uniquely written as

${\displaystyle rs^{2}}$

where r is square-free, or not divisible by any square numbers (let s2 be the largest square number that divides n and then let r = n/s2). Now suppose that there are only finitely many prime numbers and call the number of prime numbers k. As each of the prime numbers factorizes any squarefree number at most once, by the fundamental theorem of arithmetic, there are only 2k square-free numbers (see Combination#Number of k-combinations for all k).

Now fix a positive integer N and consider the integers between 1 and N. Each of these numbers can be written as rs2 where r is square-free and s2 is a square, like this:

( 1×1, 2×1, 3×1, 1×4, 5×1, 6×1, 7×1, 2×4, 1×9, 10×1, ...)

There are N different numbers in the list. Each of them is made by multiplying a squarefree number, by a square number that is N or less. There are ⌊N⌋ such square numbers. Then, we form all the possible products of all squares less than N multiplied by all squarefrees everywhere. There are exactly 2kN⌋ such numbers, all different, and they include all the numbers in our list and maybe more. Therefore, 2kN⌋ ≥ N. Here, ⌊x⌋ denotes the floor function.

Since this inequality does not hold for N sufficiently large, there must be infinitely many primes.

## Furstenberg's proof

In the 1950s, Hillel Furstenberg introduced a proof using point-set topology.

## Some recent proofs

### Pinasco

Juan Pablo Pinasco has written the following proof.[6]

Let p1, ..., pN be the smallest N primes. Then by the inclusion–exclusion principle, the number of positive integers less than or equal to x that are divisible by one of those primes is

{\displaystyle {\begin{aligned}1+\sum _{i}\left\lfloor {\frac {x}{p_{i}}}\right\rfloor -\sum _{i

Dividing by x and letting x → ∞ gives

${\displaystyle \sum _{i}{\frac {1}{p_{i}}}-\sum _{i

This can be written as

${\displaystyle 1-\prod _{i=1}^{N}\left(1-{\frac {1}{p_{i}}}\right).\qquad (3)}$

If no other primes than p1, ..., pN exist, then the expression in (1) is equal to ${\displaystyle \lfloor x\rfloor }$  and the expression in (2) is equal to 1, but clearly the expression in (3) is not equal to 1. Therefore, there must be more primes than  p1, ..., pN.

### Whang

In 2010, Junho Peter Whang published the following proof by contradiction.[7] Let k be any positive integer. Then according to de Polignac's formula (actually due to Legendre)

${\displaystyle k!=\prod _{p{\text{ prime}}}p^{f(p,k)}}$

where

${\displaystyle f(p,k)=\left\lfloor {\frac {k}{p}}\right\rfloor +\left\lfloor {\frac {k}{p^{2}}}\right\rfloor +\cdots .}$
${\displaystyle f(p,k)<{\frac {k}{p}}+{\frac {k}{p^{2}}}+\cdots ={\frac {k}{p-1}}\leq k.}$

But if only finitely many primes exist, then

${\displaystyle \lim _{k\to \infty }{\frac {\left(\prod _{p}p\right)^{k}}{k!}}=0,}$

(the numerator of the fraction would grow singly exponentially while by Stirling's approximation the denominator grows more quickly than singly exponentially), contradicting the fact that for each k the numerator is greater than or equal to the denominator.

### Saidak

Filip Saidak gave the following proof by construction, which does not use reductio ad absurdum[8] or Euclid's Lemma (that if a prime p divides ab then it must divide a or b).

Since each natural number (> 1) has at least one prime factor, and two successive numbers n and (n + 1) have no factor in common, the product n(n + 1) has more different prime factors than the number n itself.  So the chain of pronic numbers:
1×2 = 2 {2},    2×3 = 6 {2, 3},    6×7 = 42 {2,3, 7},    42×43 = 1806 {2,3,7, 43},    1806×1807 = 3263443 {2,3,7,43,13,139}, · · ·
provides a sequence of unlimited growing sets of primes.

## Proof using the irrationality of π

Representing the Leibniz formula for π as an Euler product gives[9]

${\displaystyle {\frac {\pi }{4}}={\frac {3}{4}}\times {\frac {5}{4}}\times {\frac {7}{8}}\times {\frac {11}{12}}\times {\frac {13}{12}}\times {\frac {17}{16}}\times {\frac {19}{20}}\times {\frac {23}{24}}\times {\frac {29}{28}}\times {\frac {31}{32}}\times \cdots }$

The numerators of this product are the odd prime numbers, and each denominator is the multiple of four nearest to the numerator.

If there were finitely many primes this formula would show that π is a rational number whose denominator is the product of all multiples of 4 that are one more or less than a prime number, contradicting the fact that π is irrational.

## Proof using information theory

Alexander Shen and others have presented a proof that uses incompressibility:[10]

Suppose there were only k primes (p1... pk). By the fundamental theorem of arithmetic, any positive integer n could then be represented as:

${\displaystyle n={p_{1}}^{e_{1}}{p_{2}}^{e_{2}}\cdots {p_{k}}^{e_{k}},}$

where the non-negative integer exponents ei together with the finite-sized list of primes are enough to reconstruct the number. Since ${\displaystyle p_{i}\geq 2}$  for all i, it follows that all ${\displaystyle e_{i}\leq \lg n}$  (where ${\displaystyle \lg }$  denotes the base-2 logarithm).

This yields an encoding for n of the following size (using big O notation):

${\displaystyle O({\text{prime list size}}+k\lg \lg n)=O(\lg \lg n)}$  bits.

This is a much more efficient encoding than representing n directly in binary, which takes ${\displaystyle N=O(\lg n)}$  bits. An established result in lossless data compression states that one cannot generally compress N bits of information into less than N bits. The representation above violates this by far when n is large enough since ${\displaystyle \lg \lg n=o(\lg n)}$ .

Therefore, the number of primes must not be finite.

3. ^ In general, for any integers a, b, c if ${\displaystyle a\mid b}$  and ${\displaystyle a\mid c}$ , then ${\displaystyle a\mid (b-c)}$ . For more information, see Divisibility.