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Channel: September 2009 – Annoying Precision
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Some remarks on the infinitude of primes

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I have at least four planned posts left in the series on symmetric functions, but unfortunately I’ll be very busy for the next two weeks. In the meantime, here are some thoughts on the primes.

Euclid’s proof of the infinitude of the primes is often held up as a shining example of mathematical proof. (Whether this reputation is deserved is a matter of opinion.) Euler’s proof via the zeta function is also classic. An even showier proof by Furstenburg is phrased in the language of topology. Today I’d like to share my personal favorite proof and discuss one of its possible consequences.

Special cases of Dirichlet’s theorem

Euclid’s proof generalizes to certain special cases of Dirichlet’s theorem. The general recipe is as follows. First, one finds an integer polynomial P(x) with the property that its integer values, with finitely many exceptions, are only divisible by primes in some congruence class. For example, the polynomial P(x) = x^2 + 1 is only divisible by primes congruent to 1 \bmod 4. Then one uses the following result.

Theorem: The set of primes dividing the integer values of an integer polynomial is infinite.

The standard proof of this result is a generalization of Euclid’s proof: if P(0) \neq 0 (which we can assume WLOG), we suppose that there are finitely many such primes p_1, p_2, \dots p_n and considers P(P(0) p_1 \dots p_n). After dividing out by P(0), we obtain a number coprime to p_1 \dots p_n; contradiction. (Edit, 2/21/22: As Alexander Stasinski says in the comments, we need to do a little more work here to make sure that P(P(0) p_1 \dots p_n) is not equal to 0 or \pm 1. Every way I can think of to do this is slightly inelegant but we can, for example, instead consider P(P(0) p_1 \dots p_n x), which can only be equal to 0 or \pm 1 for finitely many values of x, and pick x not equal to any such value.)

One obtains Euclid’s result with P(x) = x + 1 and Dirichlet’s theorem for primes congruent to 1 \bmod n using the cyclotomic polynomials P(x) = \Phi_n(x); I’ve written about this proof in an old post. A paper of Conrad discusses the generalization of this method to primes congruent to a \bmod n where a^2 \equiv 1 \bmod n; it is known that one can do no better by this method.

Polynomials are not important

The limitation of this method is that P(x) is required to be a polynomial, and the reason I no longer like the above proof is that the highly “algebraic” method does not suggest the following generalization at all.

Theorem: Let a_n be a strictly increasing sequence of positive integers which grows slower than 2^{ \sqrt[k]{n} } for every k (in particular, any sequence of polynomial growth rate). Then the set of primes dividing some a_n is infinite.

Proof. We prove the following more precise converse: any strictly increasing sequence of positive integers divisible by at most k distinct primes p_1, ... p_k grows at least as fast as 2^{ \sqrt[k]{n} }.

But this is almost obvious. Let \pi_P(n) denote the number of numbers of the form p_1^{e_1} ... p_k^{e_k} less than or equal to n. Clearly any such number must satisfy

\displaystyle \sum_{i=1}^{k} e_k \log p_k \le \log n

and it follows that 1 \le e_k \le \frac{\log n}{\log p_k}. In particular, e_k \le \log_2 n for every k. (This is not the strongest bound, but it’s enough.) It follows that

\pi_P(n) \le (\log_2 n)^k.

This is equivalent to the claim we wanted to prove. (Note that if a_n is required to be strictly increasing we get an honest inequality a_n \ge 2^{ \sqrt[k]{n} }; otherwise, we can still say \liminf_{n \to \infty} \frac{a_n}{2^{ \sqrt[k]{n} } } \ge 1.)

I find this proof the best motivated out of any of the proofs I know, especially because ultimately it reduces to simple counting. It is also constructive in the sense that a failure to satisfy the above inequality for some k, n (for a strictly increasing sequence) implies that you have already hit at least k prime factors.

Motivated by the above result, a positive solution to the following problem would imply a proof of Dirichlet’s theorem that totally avoids the machinery of zeta functions.

Problem: Explicitly describe a sequence a_n growing slower than 2^{ \sqrt[k]{n} } for every k that is divisible, with finitely many exceptions, only by primes in a specified arithmetic progression.

Of course we already know by the strong form of Dirichlet’s theorem that this is true of the sequence a_n consisting precisely of the numbers divisible by primes in a specified arithmetic progression. But this is a “high-complexity” sequence whereas, say, the integer values of a cyclotomic polynomial \Phi_n(x) are “low-complexity.” In the spirit of Polymath4 what would be nice is a sequence describable in polynomial time, but as long as the proof that both of the above conditions are met is elementary it really doesn’t matter. Ideally a_n would be some perturbation of a polynomial, so the first condition would be trivial to verify. I have no idea how one would go about constructing such a sequence.


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