Every Even Number, Two Primes

Choose an even whole number greater than 2. Try primes no larger than half that number: subtract each candidate from the chosen number and check whether the remainder is also prime. A prime is a whole number greater than 1 with no positive divisors except 1 and itself.

For 10, subtracting 2 leaves 8, which is not prime. Subtracting 3 leaves 7, giving 10 = 3 + 7, and subtracting 5 leaves 5, giving 10 = 5 + 5. There is no need to try 7, because it would give the already-counted pair 7 + 3. Equal primes are allowed.

The second view counts these pairs for many even starting values. Goldbach's conjecture says each even number greater than 2 has at least one pair, not that it has only one.

The value 2 is available as an exception to inspect: the smallest sum of two primes is 2 + 2 = 4, so 2 has no pair and is outside the conjecture. Each larger value can be checked individually; what remains unproved is the statement about all of them.

Once the numbers get large, a pair is not a coincidence. It is what you would expect from a great many attempts.

Below $n$ there are roughly $n/\ln n$ primes. To write $n$ as a pair you may try any prime up to $n/2$, which gives you something like $n/(2\ln n)$ attempts, and each attempt asks whether one particular number near $n$ happens to be prime, which has a chance of roughly $1/\ln n$. Multiply those together and the expected number of pairs is around $n/(2(\ln n{)}^2)$, which grows. For $n=100,000$ that is a few hundred, and the actual count is 810.

That is the whole intuition, and it explains the shape of the comet: more attempts and a growing success rate means the counts climb, and the lower edge of the spray climbs with them. It is also the reason nobody expects a counterexample. An even number with no pair at all would have to have every single one of those attempts fail.

And it is not a proof, for a reason worth being precise about. Treating "is $n-p$ prime" as a coin flip with probability $1/\ln n$ is a model of the primes, and the primes are not random. They are a completely determined set, and the conjecture is exactly the assertion that this particular determined set never conspires against one particular even number. An expected value of four hundred does not forbid an outcome of zero; it only makes it look unlikely under an assumption nobody can justify.

The rule switch on this page is there to make that concrete. Ask for both primes to be one more than a multiple of four and the same counting argument still predicts plenty of pairs, and yet every multiple of four has none, for a reason the counting argument cannot see.

The sentence worth carrying away is this one.

A search that never fails has established what it looked at, and has said nothing about what it did not.

This page checks every even number up to a million. That is 499,999 complete searches, all of them successful, and the fact established is precisely: every even number from 4 to 1,000,000 is a sum of two primes. Not "probably all even numbers". Not "all even numbers, subject to checking". The range is the claim.

Somebody else has done the same thing to 4×10^18. That is an extraordinary computation and it is the same kind of statement, eighteen orders of magnitude along. The gap between it and the conjecture is not smaller than the gap between this page and the conjecture, because both gaps are infinite.

Three things on this page are established in three different ways, and the whole point is knowing which is which.

• Ten is 3 + 7 and 5 + 5, and 2 has no pair at all. Both are proved, by finite arithmetic that is carried out in full.
• Every even number from 4 to a million has a pair. That is exhaustively checked over exactly that range, and it is offered as a fact about that range and not as evidence for anything.
• Every even number greater than 2 has a pair. That is open, it is cited, and nothing on this page bears on it.

The temptation is for the second to slide into the third. It is a strong temptation here, because the second has no exceptions and looks like a demonstration. The way to resist it is to notice that the restricted rule would look just as convincing if you only ever tried the numbers that work.

For an even number $n$, let

1. $r(2)=0$, since the least sum of two primes is $2+2=4$. Proved.
2. $r(10)=2$, with the pairs $3+7$ and $5+5$. Proved, by completing the search over primes up to $5$.
3. $r(n)\ge 1$ for every even $n$ with $4\le n\le 10^6$. Exhaustively checked over exactly that range, with the search for each $n$ stopping at the first pair found.
4. $\max \{r(n):4\le n\le 10^5\}=2168$, attained at $n=99,330$, and $\min \{r(n):4\le n\le 10^5\}=1$, attained at $n=4$. Exhaustively checked over exactly that range.
5. $r(n)\ge 1$ for every even $n>2$. Open (Goldbach and Euler, 1742). Verified computationally for $n\le 4\times 10^{18}$ by Oliveira e Silva, Herzog and Pardi, which is a search result and not a proof.
6. Every odd $n>5$ is a sum of three primes. Proved (Helfgott, 2013), cited and not reproduced here.
7. Let $r_1(n)$ count pairs with both primes congruent to $1\mathrm{mod}4$. Then $r_1(n)=0$ whenever $4\mid n$. Proved, in one line, below.
8. $r(n)\sim C_2{\prod }_{p\mid n,p>2}\frac{p-1}{p-2}\cdot \frac{n}{(\ln n{)}^2}$, where $C_2\approx 0.6601618$. Heuristic (Hardy and Littlewood, Conjecture A), drawn over the comet as a prediction.

Statement 5 is the exploration. Statement 3 is what this page can establish on its own, and it is consistent with either answer to 5 without being evidence for either.

Two has no pair. Every prime is at least 2, so every sum of two primes is at least 4. Hence $r(2)=0$. ∎

Ten has exactly two. Any pair $p+q=10$ with $p\le q$ has $p\le 5$, so $p\in \{2,3,5\}$ is the complete list of candidates. Then $10-2=8=2^3$, which is not prime; $10-3=7$, prime; $10-5=5$, prime. So $r(10)=2$, and the search over three candidates is exhaustive because the constraint $p\le q$ leaves nothing above 5 to try. ∎

The restricted rule fails on every multiple of four. Suppose $p\equiv 1(\mathrm{mod}4)$ and $q\equiv 1(\mathrm{mod}4)$. Then $p+q\equiv 2(\mathrm{mod}4)$. A multiple of 4 is congruent to $0\mathrm{mod}4$, so it is not of that form, and no such pair can exist for it. Note what this argument does not require: no primality, no searching, no bound. It is a statement about all multiples of four at once, which is precisely the kind of statement the search on this page cannot produce. ∎

Every even number below a million has a pair. Sieve to $10^6$, then for each even $n$ in range walk the primes upward until one leaves a prime behind. Each of the 499,999 searches terminates with a pair, in exact integer arithmetic. This proves statement 3 and only statement 3. ∎

What is not proved here. Statement 5, in either direction. It is worth being explicit about why no amount of computing gets closer. Every even number checked removes one candidate counterexample from an infinite supply, so the proportion remaining is unchanged, and the conjecture is not the sort of claim that becomes more likely as the checked range grows. A proof would have to be an argument about all even numbers, and the closest anyone has come is Chen's theorem of 1973, which says every sufficiently large even number is a prime plus a number with at most two prime factors. That is a genuinely different statement and it is the state of the art.

What this laboratory computes. Searches for any even number up to 100,000, with primality from a sieve of Eratosthenes in exact integer arithmetic. The comet counts every pair for every even number up to 100,000, which is about fifty thousand complete searches and runs in a background thread with a working progress bar and a cancel that really stops it. The largest rung, a million, does something different and cheaper: it stops at the first pair for each number, so what it can report is that a pair exists and how small the smallest one was. Counting every pair to a million would be roughly a hundred times the arithmetic of counting to a hundred thousand, and this page declines it rather than making you wait.

The prediction is not a fit. The curve drawn over the comet comes from Hardy and Littlewood's Conjecture A, evaluated at each even number from its own odd prime factors. Nothing about it is derived from the counts underneath it. At these sizes it runs about a fifth below them, and it would still run where it runs if every count on the figure were different, which is the property that makes it a prediction rather than a description.

Where the mathematics is still moving. The even conjecture is open. The odd one was closed by Helfgott in 2013, and the distance between those two facts is the most interesting thing on this page: the two statements are one word apart in English and very far apart in difficulty, because three primes give a circle-method argument enough room to work and two do not. Chen's theorem gets within one almost-prime. Nobody has an approach that gets the rest of the way.

Read this next to Gilbreath's difference field. That page has the same shape of problem in the same objects: a pattern in the primes that has held for every case anyone has computed and that nobody can prove. The pages differ in what the reader is asked to resist. On that one the surprise is that anything regular appears at all. Here there is no surprise in the picture, and the thing to resist is the feeling that half a million successful searches amount to an argument.