For any finite set \(\{ p_1, \dots , p_r\} \) of primes, consider the number \(n = p_1 p_2 \cdots p_r + 1\). This \(n\) has a prime divisor \(p\). But \(p\) is not one of the \(p_i\)’s: otherwise \(p\) would be a divisor of \(n\) and of the product \(p_1p_2\dots p_r\), and thus also of the difference \(n - p_1 p_2 \cdots p_r = 1\), which is impossible. So a finite set \(\{ p_1, \dots , p_r\} \) cannot be the collection of all prime numbers.

Let us first look at the Fermat numbers \(F_n = 2^{2^n} +1\) for \(n = 0,1,2,\dots \). We will show that any two Fermat numbers are relatively prime; hence there must be infinitely many primes. To this end, we verify the recursion

from which our assertion follows immediately. Indeed, if \(m\) is a divisor of, say, \(F_k\) and \(F_n\) (with \(k {\lt} n\)), then \(m\) divides 2, and hence \(m = 1\) or \(2\). But \(m = 2\) is impossible since all Fermat numbers are odd. To prove the recursion we use induction on \(n\). For \(n = 1\), we have \(F_0 = 3\) and \(F_1 - 2 = 3\). With induction we now conclude

Suppose \(\mathbb {P}\) is finite and \(p\) is the largest prime. We consider the so-called Mersenne number \(2^p - 1\) and show that any prime factor \(q\) of \(2^p - 1\) is bigger than \(p\), which will yield the desired conclusion. Let \(q\) be a prime dividing \(2^p - 1\), so we have \(2^p \equiv 1 \pmod{q}\). Since \(p\) is prime, this means that the element 2 has order \(p\) in the multiplicative group \(\mathbb {Z}_q \setminus \{ 0\} \) of the field \(\mathbb {Z}_q\). This group has \(q - 1\) elements. By Lagrange’s theorem, we know that the order of every element divides the size of the group, that is, we have \(p \mid q - 1\), and hence \(p {\lt} q\).

Let \(\pi (x) := \# \{ p \leq x : p \in \mathbb {P}\} \) be the number of primes that are less than or equal to the real number \(x\). We number the primes \(\mathbb {P} = \{ p_1, p_2, p_3, \dots \} \) in increasing order. Consider the natural logarithm \(\log x\), defined as

Now we compare the area below the graph of \(f(t) = \frac{1}{t}\) with an upper step function. (See also the appendix for this method.) Thus for \(n \leq x {\lt} n+1\) we have

where the sum extends over all \(m \in \mathbb {N}\) which have only prime divisors \(p \leq x\).

Since every such \(m\) can be written in a unique way as a product of the form \(\prod _{p \leq x} p^{k_p}\), we see that the last sum is equal to

The inner sum is a geometric series with ratio \(\frac{1}{p}\), hence

Now clearly \(p_k \geq k+1\), and thus

and therefore

Everybody knows that \(\log x\) is not bounded, so we conclude that \(\pi (x)\) is unbounded as well, and so there are infinitely many primes.

Consider the following curious topology on the set \(\mathbb {Z}\) of integers. For \(a, b \in \mathbb {Z}, b {\gt} 0\), we set

Each set \(N_{a,b}\) is a two-way infinite arithmetic progression. Now call a set \(O \subseteq \mathbb {Z}\) open if either \(O\) is empty, or if to every \(a \in O\) there exists some \(b {\gt} 0\) with \(N_{a,b} \subseteq O\). Clearly, the union of open sets is open again. If \(O_1, O_2\) are open, and \(a \in O_1 \cap O_2\) with \(N_{a,b_1} \subseteq O_1\) and \(N_{a,b_2} \subseteq O_2\), then \(a \in N_{a, b_1 b_2} \subseteq O_1 \cap O_2\). So we conclude that any finite intersection of open sets is again open. Therefore, this family of open sets induces a bona fide topology on \(\mathbb {Z}\).

Let us note two facts:

• Any nonempty open set is infinite.

• Any set \(N_{a,b}\) is closed as well.

Indeed, the first fact follows from the definition. For the second, we observe

which proves that \(N_{a,b}\) is the complement of an open set and hence closed.

So far, the primes have not yet entered the picture — but here they come. Since any number \(n \neq 1, -1\) has a prime divisor \(p\), and hence is contained in \(N_{0,p}\), we conclude

Now if \(\mathbb {P}\) were finite, then \(\bigcup _{p \in \mathbb {P}} N_{0,p}\) would be a finite union of closed sets (by (B)), and hence closed. Consequently, \(\{ 1, -1\} \) would be an open set, in violation of (A).

Our final proof goes a considerable step further and demonstrates not only that there are infinitely many primes, but also that the series \(\sum _{p \in \mathbb {P}} \frac{1}{p}\) diverges. The first proof of this important result was given by Euler (and is interesting in its own right), but our proof, devised by Erdős, is of compelling beauty.

Let \(p_1, p_2, p_3, \dots \) be the sequence of primes in increasing order, and assume that \(\sum _{p \in \mathbb {P}} \frac{1}{p}\) converges. Then there must be a natural number \(k\) such that \(\sum _{i \geq k+1} \frac{1}{p_i} {\lt} \frac{1}{2}\). Let us call \(p_1, \dots , p_k\) the small primes, and \(p_{k+1}, p_{k+2}, \dots \) the big primes. For an arbitrary natural number \(N\), we therefore find

Let \(N_b\) be the number of positive integers \(n \leq N\) which are divisible by at least one big prime, and \(N_s\) the number of positive integers \(n \leq N\) which have only small prime divisors. We are going to show that for a suitable \(N\)

which will be our desired contradiction, since by definition \(N_b + N_s\) would have to be equal to \(N\).

To estimate \(N_b\), note that \(\left\lfloor \frac{N}{p_i} \right\rfloor \) counts the positive integers \(n \leq N\) which are multiples of \(p_i\). Hence by (1) we obtain

Let us now look at \(N_s\). We write every \(n \leq N\) which has only small prime divisors in the form \(n = a_n b_n^2\), where \(a_n\) is the square-free part. Every \(a_n\) is thus a product of different small primes, and we conclude that there are precisely \(2^k\) different square-free parts. Furthermore, as \(b_n^2 \leq n \leq N\), we find that there are at most \(\sqrt{N}\) different square parts, and so

Since (2) holds for any \(N\), it remains to find a number \(N\) with \(2^k \sqrt{N} {\lt} \frac{N}{2}\), or \(2^{k+1} {\lt} \sqrt{N}\), and for this \(N = 2^{2k+2}\) will do.

We may assume that \(f(n)\) is monotonely increasing. Otherwise, replace \(f(n)\) by \(F(n) = \max _{i \leq n} f(i)\); you can easily check that with this \(F(n)\) the sequence \(S\) again satisfies the subexponential growth condition.

Let us suppose for a contradiction that \(\mathbb {P}_S = \{ p_1, \dots , p_k\} \) is finite. For \(n \in \mathbb {N}\), let

where the \(\alpha _i = \alpha _i(n)\) depend on \(n\). (For \(s_n = 0\) we can put \(\alpha _i = 0\) for all \(i\).) Then

and thus by taking the binary logarithm

Hence there are not more than \(2^{f(n)} + 1\) different possible values for each \(\alpha _i = \alpha _i(n)\). Since \(f\) is monotone, this gives a first estimate

On the other hand, since \(S\) is almost injective only \(c\) terms in the sequence can be equal to 0, and each nonzero absolute value can occur at most \(2c\) times, so we get the lower estimate

Altogether, this gives

Taking again the logarithm with base 2 on both sides, we obtain

This, however, is plainly false for large \(N\), as \(k\) and \(c\) are constants, so \(\frac{\log _2(N-c)}{\log _2 N}\) goes to 1 for \(N \to \infty \), while \(\frac{f(N)}{\log _2 N}\) goes to 0.

See theorems in this chapter.