Euclid's theorem

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.[8]

Juan Pablo Pinasco has written the following proof.[11]

Let p₁, ..., p_N 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

${\displaystyle {\begin{aligned}1+\sum _{i}\left\lfloor {\frac {x}{p_{i}}}\right\rfloor -\sum _{i<j}\left\lfloor {\frac {x}{p_{i}p_{j}}}\right\rfloor &+\sum _{i<j<k}\left\lfloor {\frac {x}{p_{i}p_{j}p_{k}}}\right\rfloor -\cdots \\&\cdots \pm (-1)^{N+1}\left\lfloor {\frac {x}{p_{1}\cdots p_{N}}}\right\rfloor .\qquad (1)\end{aligned}}}$

Dividing by x and letting x → ∞ gives

${\displaystyle \sum _{i}{\frac {1}{p_{i}}}-\sum _{i<j}{\frac {1}{p_{i}p_{j}}}+\sum _{i<j<k}{\frac {1}{p_{i}p_{j}p_{k}}}-\cdots \pm (-1)^{N+1}{\frac {1}{p_{1}\cdots p_{N}}}.\qquad (2)}$

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, ..., p_N 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, ..., p_N.

In 2010, Junho Peter Whang published the following proof by contradiction.[12] 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.

Filip Saidak gave the following proof by construction, which does not use reductio ad absurdum[13] 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 = 3263442 {2, 3, 7, 43, 13, 139}, · · ·
provides a sequence of unlimited growing sets of primes.

Representing the Leibniz formula for π as an Euler product gives[14]

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

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.

Alexander Shen and others have presented a proof that uses incompressibility:[15]

Suppose there were only k primes (p₁... p_k). 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 e_i 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 fewer 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.

Dirichlet's theorem states that for any two positive coprime integers a and d, there are infinitely many primes of the form a + nd, where n is also a positive integer. In other words, there are infinitely many primes that are congruent to a modulo d.

Let π(x) be the prime-counting function that gives the number of primes less than or equal to x, for any real number x. The prime number theorem then states that x / log x is a good approximation to π(x), in the sense that the limit of the quotient of the two functions π(x) and x / log x as x increases without bound is 1:

${\displaystyle \lim _{x\rightarrow \infty }{\frac {\pi (x)}{x/\ln(x)}}=1.}$

Using asymptotic notation this result can be restated as

${\displaystyle \pi (x)\sim {\frac {x}{\log x}}.}$

This yields Euclid's theorem, since ${\displaystyle \lim _{x\rightarrow \infty }{\frac {x}{\log x}}=\infty .}$