Every large even number is a prime plus an almost-prime

The question

In 1742 Christian Goldbach guessed that every even number from 4 upward can be written as a sum of two primes: 10 = 3 + 7, 100 = 47 + 53, and so on. Nobody has found an exception, and nobody has proved it.

Chen proved the closest thing we have. He allowed the second number to be slightly less than prime: either a prime, or a product of exactly two primes (such as 15 = 3 × 5 or 49 = 7 × 7). Mathematicians call that an almost-prime.

Try it. Pick any even number and the panel finds every way to write it in both styles.

The shorthand for this result is “1 + 2”. It has nothing to do with 1 + 2 = 3. It means: one number with 1 prime factor, plus one number with at most 2 prime factors. In that shorthand, Goldbach's own conjecture is “1 + 1”. The paper writes it as the proposition (1, 2).

The ladder to 1+2

Nobody jumped straight to 1+2. For fifty years people proved weaker statements “a + b”: every large even number is a sum of one number with at most a prime factors and another with at most b. Smaller is better. Click a rung.

The step from “both numbers are almost-primes” to “one of them is a genuine prime” (Rényi, 1948) was a big change, because primes are much harder to control than almost-primes. After that the game was to push the second number down: some fixed c, then 5, 4, 3. Chen announced 1+2 in 1966 with only a sketch. This 1973 paper is the full proof, and it also improves his constant from 0.098 to 0.67 (Chapter 3 explains what that constant is).

What exactly is proved

Chen does more than show one representation exists. He shows there are many, and gives a guaranteed minimum. This is Theorem I, with x the even number:

P_x(1,2)

The count from Chapter 1: how many primes p make x − p a prime or a product of two primes.

x / (log x)²

The natural size of the answer. Roughly x / log x of the numbers below x are prime, and each x − p then has roughly a 1-in-log-x chance of being prime-like. Multiply the two. (“log” is the natural logarithm throughout.)

C_x

A correction that depends on which small primes divide x. It starts at about 0.66 and is multiplied by (q−1)/(q−2) for each odd prime q dividing x. Multiples of 3 get double.

0.67

The number the whole paper exists to produce. It only has to be bigger than zero. Any positive constant means the count grows without limit, so it can never be zero.

Why would dividing by 3 help? If 3 divides x, then x − p is never a multiple of 3 (unless p = 3), so it is more likely to be prime-like. You can see this as two separate bands in the real data:

The catch: “sufficiently large”

The theorem holds for all even numbers beyond some threshold. The paper does not say where the threshold is, and the method cannot give a sensible one: several steps only work once log x is enormous (Lemma 1 already asks for log x ≥ 1000, so x has more than 400 digits). So Chen's theorem does not settle any particular even number. It settles all of them from some point on.

Theorem II: a twin for twin primes

The same machine proves a second result. Fix any even gap h. Then there are infinitely many primes p such that p + h is a prime or a product of two primes, with the same kind of lower bound. With h = 2 this is the nearest known thing to the twin prime conjecture.

Sieves, and why they stall

The basic tool is a sieve. You know the oldest one: to find primes, write down the numbers and cross out the multiples of 2, then of 3, then of 5. What survives has no small prime factor.

For Goldbach we sieve a special list. Fix the even number x. For each prime p below x, write down the leftover x − p. We want leftovers with very few prime factors. So cross out every leftover that has a prime factor below some cutoff z.

So the cutoff x^1/10 leaves numbers with at most 9 prime factors. The cutoff x^1/3 leaves at most 2, which is exactly what we want. Drag the slider and watch the survivors.

Why not just set the cutoff to x^1/3?

In the panel you can, because the computer checks every number. A proof for all large x cannot check anything. It has to estimate how many survive, and the estimate needs one piece of input for each modulus d:

How many leftovers x − p are divisible by d? Equivalently: how many primes p leave the same remainder as x when divided by d?

The expected answer is “an equal share”: primes should spread evenly over the allowed remainders. Look for yourself.

A sieve's strength depends on how large d can get while we can still prove the spread is even. The best tool available to Chen was Bombieri's theorem (1965, reference [9] in the paper): on average, the spread is even for all d up to about √x. People say the primes have “level of distribution ½”.

With level ½, the best pure sieves gave 1+3. To reach 1+2 by sieving alone you would need more than Bombieri's theorem gives. Chen's insight was to stop trying to sieve harder, and to change what is being counted.

Chen's scoring trick

Chen sieves only lightly, with cutoff x^1/10. Too many leftovers survive, including bad ones with 3 to 9 prime factors. Then he gives each survivor a score, designed so that bad survivors score zero or less. If the total score is positive, some good survivor must exist.

Sort the prime factors of a surviving leftover into two sizes:

medium between x^1/10 and x^1/3  ·  big above x^1/3

There can never be three big ones (their product would exceed x). The score is:

Build a leftover below and see what it scores.

Run through the cases and you find the pattern that makes the proof work:

The only dangerous bad shape is medium × big × big. A plain “minus ½ per medium factor” still leaves it with ½. That is why Chen adds the special extra penalty for exactly that shape. Counting how often that shape occurs is the new problem he had to solve, and it is where most of the paper's pages go.

In the paper this is the single line (34) at the start of Section III, described there as “obvious”:

A

How many leftovers survive the light sieve. One point each. (The paper writes P_x(x, x^1/10).)

B

How many medium prime factors the survivors have in total. Half a point off for each.

Ω

How many survivors have the shape medium × big × big. Half a point off for each.

x^0.91

A negligible allowance for leftovers with a repeated prime factor, which the neat case table above glosses over.

The scoring rule is pure logic, so it is true for every even number, not only huge ones. Here is the ledger for a number you choose.

The logic is done. What remains is three estimates, valid for every large x: a floor for A − ½B and a ceiling for Ω. Measured in units of x·C_x / (log x)², Chen gets:

The gain: A − ½B is at least 2.6408

Both A and B are ordinary sieve questions about the leftovers x − p, so Chen uses tools that already existed: Bombieri's theorem for the “equal share” input, and a sharp sieve formula from Richert's 1969 paper (reference [11]).

The sieve formula says: the true number of survivors is trapped between two multiples of the naive guess,

where u measures how modest the sieve is:

The larger u, the tighter the trap. At u = 5 the two bounds almost touch, which is why Chen chose a cutoff as low as x^1/10. A light sieve can be estimated almost exactly.

How the two pieces are estimated

A (a floor). Use the lower curve at u = 5. This is formula (31). In our units it gives 8 × (log 4 + a small integral).

B (a ceiling). For each medium prime q, count survivors divisible by q. That is again a sieve, but part of the information budget is used up by q itself, so u drops from 5 to somewhere between 3 and 4. Use the upper curve there, and add up over all medium q. This is formula (32).

Subtract. The two small integrals partly cancel. Chen bounds what is left by hand, in formula (33), using only an elementary inequality for the logarithm:

The page recomputes the leftover integral exactly: it is about –, a little better than Chen's safe hand estimate of −0.0164725. His arithmetic errs on the safe side, as a proof must.

The loss: Ω is at most 3.9404

Ω counts primes p for which x − p = p₁p₂p₃. The sieve on the leftovers cannot see this shape. Here is Chen's new idea.

The roles have swapped. Before, the list was indexed by primes p and we looked for almost-prime leftovers. Now the list is indexed by products of three primes and we look for prime leftovers. An upper bound for primes in a list is the easiest job a sieve does.

What the products look like

Think on a logarithmic scale, where multiplying becomes adding lengths. A product near x is a bar of length 1 cut into three pieces: p₁ = x^α, p₂ = x^β, and p₃ takes the rest.

The rules in the paper are 1/10 < α ≤ 1/3 < β ≤ (1 − α)/2. The last one just says p₂ is the smaller of the two big primes.

The upper-bound sieve, in three moves

Move 1: Selberg's squaring trick. Chen picks numbers λ_d, one for each d up to about x^1/4, with λ₁ = 1. For any number n, add up the λ_d over the small divisors d of n and square the result. If n has no small prime factors, the only divisor is 1 and the square is exactly 1. Otherwise the square is still at least 0. So the sum of these squares over a list is always at least the number of sieve survivors, whatever the λ_d are. Selberg worked out the choice that makes the sum smallest. The paper defines these λ_d just after Lemma 4.

Move 2: split into main part and error. Expanding the square gives a main term M₁ (what you get if the products spread perfectly evenly over remainders) and an error term M₂ (the total unevenness). This is Lemma 5: Ω ≤ M₁ + M₂, up to negligible terms.

Move 3: estimate each. Lemma 6 shows M₂ is negligible. That is the hard part, covered in Chapter 8. Lemma 7 evaluates M₁:

The factor 8 is the price of the sieve. The truth is believed to be about 2 in place of 8, so the method overshoots roughly fourfold. Chen's scoring scheme has enough slack to absorb that.

Counting the products

The number of products depends only on the allowed cuts (α, β). It comes out as x / log x times an integral over the allowed region, which Chen bounds by hand in formula (28), splitting the range into seven strips and using logarithm tables:

The page evaluates the same integral numerically and gets –. Again Chen's hand bound is slightly on the safe side. The abstract's phrase “without any complicated numerical calculations” refers to this: one page of arithmetic you can check with a pocket calculator.

The engine room

Move 2 above needed one input: the products p₁p₂p₃ must spread evenly over remainders, on average, for all moduli d up to about √x. Bombieri's theorem says this for primes. Nobody had proved it for Chen's products, so he proved it himself. That is Lemma 6, and Lemmas 1 to 4 are its tools. Go back to the remainder panel in Chapter 4 and press “Chen's products” to see the claim on real numbers.

This is the most technical half of the paper. You do not need to follow each line to understand the proof's structure. Here is what each tool does.

Lemma 1: a smooth on-switch

The count includes only products below x. A sharp cutoff is awkward for the calculus that follows, so Chen replaces it by a function Φ that is 0 up to the cutoff, rises smoothly, and is 1 soon after. He writes it as a contour integral. Evaluated, it is a familiar object: the chance that a Poisson random count exceeds log x.

Lemma 2: the large sieve

To test whether a list favours some remainder mod d, you multiply it against test patterns that repeat every d steps. These patterns are called characters, written χ. A list that is evenly spread gives a small answer against every pattern.

The large sieve inequality says a list cannot correlate strongly with many different patterns at once. Add the squared correlations over all patterns for all d up to Q, and the total is at most about (Q² + length of list) × (size of list). It is a budget: a little unevenness in total, to be shared among all moduli. So the average modulus gets almost none.

Lemma 3 and Lemma 4

Lemma 3 applies the large sieve to L-functions. An L-function packages a character into a function of a complex variable, the way the Riemann zeta function packages the plain integers. The lemma bounds their size on average (a fourth-power average). Lemma 4 is a small bookkeeping bound on sums of characters, used once to handle prime powers.

Lemma 6: putting the tools to work

Why can Chen do for p₁p₂p₃ what is so hard for other lists? Because a product has two independent parts: the pair p₁p₂ and the third prime. A correlation of the product with a pattern factors into a correlation of each part, and the large sieve handles each part separately. A list without this product structure gives the method nothing to hold.

The proof splits the moduli d into three ranges:

In every range the unevenness is smaller than the main term by a power of log x. Result: M₂ ≪ x / (log x)^2.01. The exponent 2.01 is what matters. The main terms have (log x)² below, so for large x the error is too small to affect the constant.

Adding it up

The paper's final section is five lines long. Put the two estimates into the scoring inequality:

The total score is positive, so good leftovers exist, and Theorem I is proved. The margin is real but not large. Slide the Ω bound to see how much worse it could have been.

The break-even point is Ω ≤ 5.2816. Chen's 3.9404 clears it comfortably. If the upper-bound sieve had cost a factor 11 instead of 8, the proof would fail. For Theorem II, the paper says to run the same argument with p + h in place of x − p.

The whole proof on one screen

Reading guide

Page by page

A good order for a first reading of the original: pages 287–288, then jump to the end (305), then Lemma 9, then Lemma 8, and only then the middle.

Symbols you will meet

log x

Natural logarithm. Roughly 2.3 × the number of digits of x.

p | n

“p divides n”.

(a, b)

Greatest common divisor. (a, b) = 1 means no shared factor.

A ≪ B

“A is at most some fixed constant times B.” The constant is not tracked.

O(…)

An error term of at most that size, up to a constant.

ε

A positive number as small as you like, fixed in advance.

φ(d)

Euler's function: how many remainders mod d share no factor with d. Primes have φ(d) allowed remainders to spread over.

μ(d)

Möbius function: 0 if d has a repeated prime factor, otherwise +1 or −1 by the parity of the number of prime factors.

Λ(n)

von Mangoldt function: log p if n is a power of the prime p, otherwise 0. A way of counting primes with weights that makes formulas cleaner.

χ, Σ*

A character (a repeating test pattern mod d). The star means the sum runs over “primitive” characters, those that genuinely have period d.

L(s, χ)

The L-function of χ. L′/L is its logarithmic derivative, which encodes primes weighted by χ.

e(α)

e^2πiα, a point on the unit circle.

What Chen's theorem leaves open

Goldbach's conjecture itself, 1+1, is still unproved. Chen's method cannot be pushed to reach it.