Chen’s Theorem: statement and introduction to the proof


Chen’s Theorem is one of the theorems most similar to Goldbach’s Conjecture known so far. As we said in Some important results – Chen and Yamada’s theorems, it is the work of the Chinese mathematician Chen Jingrun, who published the proof in 1966. The best known form of the statement is the following:

Chen’s Theorem

Every sufficiently large even number is the sum of either two primes, or a prime and a semiprime (that is, a product of two primes).

In other words, using the letters p, q and q^{\prime} to indicate prime numbers (not necessarily distinct), the Theorem tells us that every sufficiently large even number N can be written either in the form N = p + q or in the form N = p + qq^{\prime}. There are therefore two differences from Goldbach’s Conjecture:

  • Validity for “large enough” numbers, rather than for any even number greater than 2
  • The presence of the case N = p + qq^{\prime}

The important difference is the second one, because it leaves the doubt that there exist some even numbers that can be written in the form N = p + qq^{\prime} but not in the form N = p + q, and the statement doesn’t say how many there might be. If we succeeded in eliminating the case N = p + qq^{\prime}, and thus only the first difference remained, we would obtain a statement identical to the Conjecture up to the “large enough” number (Every sufficiently large even number is the sum of two prime numbers). This number could also be very large, even unknown, but it would still be a finite number: only a finite quantity of even numbers would be excluded from the statement, which is very little compared to all the other infinite even numbers. In mathematical terms we would say that the form N = p + q would hold for almost all even numbers, meaning all except a finite quantity. Therefore we can consider the first difference to be insignificant in the context of the infinity of even numbers, while the real limitation of Chen’s Theorem is the second difference, i.e. the presence of the case N = p + qq^{\prime}, which instead could be valid for infinite cases, given that the statement says nothing about it.

On our site we often indicate the starting even number of Goldbach’s Conjecture as 2n, to remember that it is an even number. In this path dedicated to Chen’s Theorem we will simply indicate it with the letter N. This is because we will use this number several times within complex expressions, and if we indicated it with 2n our formulas would be full of 2. Therefore in this path we must remember that N is always even (when this aspect is particularly important, we will repeat it).

The original form

After analyzing the differences with respect to Goldbach’s Conjecture, let’s go back to the statement of Chen’s Theorem. First of all we need to know that what we have seen is the abbreviated and best known form of the Theorem. Actually Chen proved the following more specific statement:

Chen’s Theorem, original version

Let N be a sufficiently large even number and let \mathfrak{R}(N) be the number of representations of N as the sum of an odd prime and a product of at most two primes. So:

\mathfrak{R}(N) \gg \mathfrak{S}(N) \frac{N}{\log^2 N} \tag{1}


\mathfrak{S}(N) = 2 \prod_{\substack{p \gt 2 \\ p \mid N}} \frac{p - 1}{p - 2} \prod_{p \gt 2} \left (1 - \frac{1}{(p - 1)^2} \right) \tag{2}

Before studying the statement in depth, two notes on the symbols used are useful:

  • The symbol \gg is just an alternative for the most commonly used “big Oh” notation; in this case, however, the asymptotic relationship “big Oh” has to be read starting from the function located towards the tip of the symbol, so (1) is equivalent to:

    \mathfrak{S}(N) \frac{N}{\log^2 N} = O(\mathfrak{R}(N))
  • The symbols \mathfrak{R} and \mathfrak{S} are the gothic versions, or more precisely Fraktur, respectively of the letters R and S. They can be read respectively as “Gothic R” and “Gothic S” and these are also frequently used notations for Chen’s Theorem or related topics.

Now let’s analyze the statement. First, it defines \mathfrak{R}(N) as the number of representations of N as the sum of an odd prime and a product of at most two primes. This means that we are considering three cases:

  • N = p + 1 (a prime plus a product of zero primes)
  • N = p + q (a prime plus another prime)
  • N = p + qq^{\prime} (a prime plus a product of two primes)

The first case is often overlooked because it can occur at most once (a number N is either of the form p + 1 or is not); the other two cases can occur several times because there is more than one variable involved, so different values of p can be combined with different values of q and/or q^{\prime}.
Furthermore, as we will see during the proof, the sums that fall within the count of \mathfrak{R}(N) in the second case are to be considered ordered, so N = q + p is considered a sum different from N = p + q if p \neq q. The reason is that in N = p + q we can alternatively consider p prime and q product of at most two primes, or vice versa; instead in the cases N = p + 1 and N = p + qq^{\prime} there is no ambiguity because 1 and qq ^{\prime} are not prime, so the prime can only be p. However, this detail is not important because, since the aim is to prove that there is at least one sum, it does not matter whether it is considered in an orderly manner or not.

Let’s calculate \mathfrak{R}(10) and \mathfrak{R}(12).

Number 10 has:

  • 0 representations as 10 = p + 1, because 9 is not prime;
  • 3 representations as 10 = p + q, because 10 = 3 + 7 = 5 + 5 = 7 + 3;
  • 0 representations such as 10 = p + qq ^{\prime}, because p can be 2, 3, 5 or 7 and in such cases, respectively, qq ^{\prime} should be 8, 7, 5 and 3 respectively, but neither of these numbers is the product of two primes (8 = 2^3 is the product of three prime numbers, while the others are prime numbers, therefore the product of only one prime).

Then \mathfrak{R}(10) = 3.

Number 12 has:

  • 1 representation as 12 = p + 1, because 11 is prime;
  • 2 representations as 12 = p + q, because 12 = 5 + 7 = 7 + 5;
  • 2 representations such as 12 = p + qq^{\prime}, because 12 = 2 + 2 \cdot 5 = 3 + 3 \cdot 3, while for p = 5, 7, 11 we’ll obtain the previous cases.

Then \mathfrak{R}(12) = 5.

It is therefore evident that Chen not only proved that every sufficiently large even number can be written as N = p + q or as N = p + qq^{\prime}, but he also tried to count how many ways it could be done (adding the case N = p + 1, which however makes little difference, since it can only happen once). However, the count of these representations is not exact, but is estimated as an asymptotic order, and the estimate is the function \mathfrak{S}(N) \frac{N}{\log^2 N}, where \mathfrak{S}(N) is given by (2).
It is convenient to rewrite \mathfrak{S}(N) as the product of two factors:

\mathfrak{S}(N) = \left(2 \prod_{p \gt 2} \left(1 - \frac{1}{(p - 1)^2} \right) \right) \left (\prod_{\substack{p \gt 2 \\ p \mid N}} \frac{p - 1}{p - 2} \right) \tag{2'}

Let’s analyze the two factors individually:

  • The factor 2 \prod_{p \gt 2} \left(1 - \frac{1}{(p - 1)^2} \right) is constant because, as we will see later, the product
    \begin{aligned} \prod_{p \gt 2} \left(1 - \frac{1}{(p - 1)^2} \right) & = \\ \left(1 - \frac{1}{2^2} \right) \left(1 - \frac{1}{4^2} \right) \left(1 - \frac{1}{6^2} \right) \left(1 - \frac{1}{10^2} \right) \ldots & = \\ \frac{3}{4} \frac{15}{16} \frac{35}{36} \frac{99}{100} \ldots \end{aligned}

    is convergent, therefore it is equal to a certain real constant (which, incidentally, is the twin prime constant, which appears in an attempt to explain another famous conjecture).

  • The factor \prod_{\substack{p \gt 2 \\ p \mid N}} \frac{p - 1}{p - 2} instead depends on N. To calculate it, in practice, a fraction of the type \frac{p - 1}{p - 2} is multiplied by every odd prime p which divides N. For example, for N = 18 = 2 \cdot 3^2 this factor is equal to \frac{3 - 1}{3 - 2} = 2 (we have to consider neither 2, nor the repetition of 3), while for N = 30 = 2 \cdot 3 \cdot 5 it’s equal to \frac{3 - 1}{3 - 2} \frac{5 - 1}{5 - 2}= 2 \cdot \frac{4}{3}.

The factor \prod_{\substack{p \gt 2 \\ p \mid N}} \frac{p - 1}{p - 2} which is present inside the he expression of \mathfrak{S}(N), and which therefore affects \mathfrak{R}(N), is a product of fractions greater than 1. Therefore, the more distinct prime divisors N has, the more fractions of this type will be multiplied together, so the bigger the result will be. Chen’s Theorem therefore seems to say that, the more distinct prime factors N has, the more possible representations of the type N = p + q or N = p + qq^{\prime} it will have. The word “seems” is a must because the count of the representations is only estimated as big Oh and not calculated exactly, however it is interesting to note that the same property holds, for the numbers verifiable by us, for Goldbach pairs (i.e. considering only sums of the type p + q). This property is more evident for the so-called “primorials”, i.e. the products of the first prime numbers: 6 = 2 \cdot 3, 30 = 2 \cdot 3 \cdot 5, 210 = 2 \cdot 3 \cdot 5 \cdot 7, 2310 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11, etc.; in fact these are the numbers that have most distinct prime factors with respect to the size of the number itself (if there were repeated prime factors, the number would increase but the distinct prime factors would stay the same, decreasing the ratio between the number of distinct prime factors and the size of the number). For these numbers, as it can be checked using the Goldbach pairs viewer, there are more Goldbach pairs than the “near” even numbers (with the only exception of 6, which is a primorial but has not more Goldbach pairs than 4 and 8; however, from 30 onwards the property is always verified and becomes more and more evident as the numbers considered increase).

How can we pass from the non-abbreviated form to the abbreviated one?

We said that Theorem C.1 is the abbreviated form of Theorem C.2. The reason lies in the fact that the second factor of (2′) is greater than 1, because it is a product of fractions greater than 1. This implies that:

\mathfrak{S}(N) \geq 2 \prod_{p \gt 2} \left(1 - \frac{1}{(p - 1)^2} \right)

But we said that the expression on the right is a constant, so \mathfrak{S}(N) is greater than a constant. With the asymptotic notation we can therefore write:

\mathfrak{S}(N) \gg 1

hence, by (1):

\mathfrak{R}(N) \gg \frac{N}{\log^2 N} \tag{3}

The last passage can be justified as follows. By (1), \mathfrak{R}(N) is definitely greater than A \cdot \mathfrak{S}(N) \frac{N}{\log^2 N}, for some positive constant A, but we said that also \mathfrak{S}(N) is greater (therefore also definitively greater) than another positive constant, which we can call B. So \mathfrak{R}(N) is definitely greater than AB \frac{N}{\log^2 N}; so, being the product AB in turn constant and positive, \mathfrak{R}(N) \gg \frac{N}{\log^2 N}.
From now on, for the sake of brevity but also so as not to bore our readers, we will no longer propose explanations of this kind in this path. If you are not familiar with this type of reasoning, before continuing with the path, we suggest that you carefully read the article Properties of asymptotic orders, where many similar properties are proved. It may also be useful to have already followed the path about the prime number Theorem, where the properties of asymptotic orders are applied frequently (especially in the last articles of the path) and their application is explained in detail.

Relation (3) means that \mathfrak{R}(N) is definitely greater than A \frac{N}{\log^2 N}, where A is a positive constant. But this function tends to infinity, so we can choose any integer M and be sure that, for N large enough, A \frac{N}{\log^2 N} \geq M; therefore, by transitivity, also \mathfrak{R}(N) is greater than or equal to M for N large enough (remember that the term “definitely” is equivalent to “for N large enough”, it’s just more generic). So we can put for example M = 2 and be certain that, for N large enough, N has at least two representations of one of the types mentioned in precedence, i.e.:

  • N = p + 1
  • N = p + q
  • N = p + qq^{\prime}

But we have observed that the first case can occur at most once; thus, if there are in total at least two representations of these types, at least one of them must be of the second or third type. Then we can say that, for N large enough, N admits at least one representation of one of two types:

  • N = p + q
  • N = p + qq^{\prime}

which is exactly what Theorem C.1 asserts.
Theorem C.2 therefore implies Theorem C.1. We can say that it is a stronger version of it, because it doesn’t just establish the existence of a representation, but, as we have seen, it says much more.

The proof

Before going into the details of the proof of Chen’s Theorem, which we will do in the next articles, it is good to have an idea of what awaits us. The first aspect to clarify is that Chen did not start from scratch but, as it often happens in research, he based his proof on theorems already proved by other mathematicians before him. More precisely, in addition to using specific theorems, he relied on an entire mathematical theory, called sieve theory, which had developed in the years prior to his 1966 proof. We could represent the main results on which Chen based his proof as in Figure 1:

Figure 1: Supporting theorems and theories for Chen’s Theorem proof

In this path we will see in detail the theory of sieves, including Jurkat-Richert’s Theorem and the so-called “large sieve inequality”, while we will deal only with the statement of the Siegel-Walfisz Theorem and the Bombieri-Vinogradov Theorem. Each of the latter, in fact, would require a separate path to be proved, as we did for the prime number Theorem.

Even if Chen’s Theorem is based on important theorems such as the ones listed, each of which has a complex proof in itself, it is too no less: in fact, many logical and algebraic steps are needed before reaching the conclusion. Despite this, there are two positive aspects of the proof:

  • It’s well structured: the initial problem, which consists in counting the pairs, is decomposed into three subproblems, each of which is addressed separately, and at the end the solutions of the three subproblems are put together to obtain an estimate of the total number of pairs, i.e. Theorem C.2.
  • It’s “elementary”, in the sense that it does not require knowledge of complex analysis, thus reducing the prerequisites required to understand it. In our site we assume only the knowledge of school mathematics and the basics of real mathematical analysis, and the proof of Chen’s Theorem fits well with this choice.

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