In 1742 Christian Goldbach wrote to Euler with a guess so simple a child can state it: every even number past two is a sum of two primes — 4 = 2+2, 6 = 3+3, 8 = 3+5, and on forever. Nearly three centuries later that binary Goldbach conjecture is still open. But its close cousin — the ternary or three-primes version, that every odd number is a sum of three primes — has been proved. That is the theorem on this page, and the story of how it fell is the story of the single most powerful machine in additive number theory: the Hardy–Littlewood circle method.

Ivan Vinogradov proved in 1937 that every sufficiently large odd number is a sum of three primes. His breakthrough was not the circle method itself — Hardy and Littlewood had built that in the 1920s — but a revolutionary new way to estimate exponential sums running over the primes, which finally supplied the one ingredient the method had been missing. Along the way we will see exactly why "three" works where "two" does not, meet the singular series that quietly explains the whole even/odd split in Goldbach's conjecture, and finish with Harald Helfgott's 2013 tour de force that removed the words "sufficiently large" and settled the ternary problem for every odd N \ge 7.

The statement

Read the formula as a story in two parts. The N^2/(\log N)^3 is the analytic or "expected" count: there are about N/\log N primes up to N by the Prime Number Theorem, so if you choose three of them to sum to a fixed N you have roughly two free choices, each carrying a 1/\log N "chance of being prime" — giving N^2/(\log N)^3. The \mathfrak{S}(N) in front is the arithmetic correction: it records how divisibility conditions (parity, being a multiple of 3, and so on) nudge the true count above or below that naive guess. Everything subtle about Goldbach lives inside that single factor.

The singular series, and why parity decides everything

The singular series is an infinite product over all primes q, of the shape

The precise algebra matters less than one structural fact: each prime q contributes a local density — the probability, computed just modulo q, that three primes can sum to N. The full product multiplies these independent local chances together, exactly in the spirit of the circle method's philosophy: a global count factors into local ones. If any single local factor vanishes, the whole product collapses to zero and there are essentially no representations.

The decisive prime is q = 2. Work modulo 2: every odd prime is \equiv 1 \pmod 2. Three odd primes therefore sum to 1 + 1 + 1 \equiv 1 \pmod 2 — an odd number. So the congruence p_1 + p_2 + p_3 \equiv N \pmod 2 can only be met when N is odd. For even N the local factor at 2 vanishes, so \mathfrak{S}(N) = 0 and three primes simply cannot reach an even target (but for the negligible use of the single even prime 2).

Worked example: watching the parity factor switch off

Let's make the vanishing concrete by counting representations modulo 2 only — a toy version of the local factor at q = 2. Take three odd primes and ask what residue their sum can land on.

Odd target, e.g. N = 21. We want p_1 + p_2 + p_3 = 21. Modulo 2 the requirement is 1 + 1 + 1 \equiv 1 \pmod 2, and 21 \equiv 1 — the residues match, the local factor is positive, and indeed the sum is realisable in many ways:

Even target, e.g. N = 20. Now we need p_1 + p_2 + p_3 = 20. With three odd primes the sum is odd, but 20 is even — a flat contradiction modulo 2. The only escape is to spend the lone even prime 2, e.g. 20 = 2 + 5 + 13. But using 2 pins one summand down completely, so it no longer behaves like a free "third prime": the count drops from an N^2/(\log N)^3 flood to a mere \sim N/(\log N)^2 trickle. That collapse is precisely \mathfrak{S}(\text{even}) = 0 speaking.

The circle method: turning a count into an integral

Here is the engine. Define the exponential sum over primes — writing e(x) = e^{2\pi i x} throughout, and weighting each prime by \log p to smooth things out:

The one magic fact about e(\cdot) is that it detects an integer:

Cube the sum and integrate. Expanding F(\alpha)^3 produces a term (\log p_1)(\log p_2)(\log p_3)\,e\big((p_1+p_2+p_3)\alpha\big) for every triple of primes, and \int_0^1 keeps exactly those triples with p_1 + p_2 + p_3 = N:

The right-hand side is our weighted representation count — positive precisely when N is a sum of three primes. So proving the theorem reduces to estimating one integral over the circle \mathbb{R}/\mathbb{Z}. The whole game is now to understand F(\alpha): where is it large, and where is it small?

Major arcs and minor arcs

The sum F(\alpha) is a battle of about N/\log N unit vectors e(p\alpha). When \alpha sits very close to a rational a/q with a small denominator q, the primes fall into arithmetic progressions modulo q and the vectors line up coherently — F(\alpha) spikes up to size nearly N/\phi(q). These neighbourhoods are the major arcs \mathfrak{M}. Everywhere else — near rationals with large denominators, or near irrationals — the vectors point every which way and mostly cancel; that leftover is the minor arcs \mathfrak{m}.

Major arcs — the main term. On \mathfrak{M} we know F(\alpha) extremely well, because the distribution of primes in arithmetic progressions modulo a small q is controlled by the Siegel–Walfisz theorem (the Prime Number Theorem in progressions, with a good uniform error). Integrating the resulting approximation over the major arcs reproduces exactly the main term \tfrac{1}{2}\mathfrak{S}(N)\,N^2/(\log N)^3 — the singular series \mathfrak{S}(N) literally is the sum of the local contributions from each major arc a/q.

Minor arcs — the enemy. Everything now rests on showing the minor-arc integral is smaller than the main term. And that is precisely where the circle method had stalled for a decade: nobody could prove F(\alpha) — a sum over the primes, an utterly irregular set — was small on \mathfrak{m}.

Vinogradov's revolution: sums over primes are small

The obstruction is that the primes have no simple formula. Hardy and Littlewood could only bound the minor arcs by assuming the Generalised Riemann Hypothesis. Vinogradov's 1937 achievement was to bound them unconditionally. His key estimate:

How do you estimate a sum over such an irregular set? Vinogradov's idea — cleanest today through Vaughan's identity — is to decompose the von Mangoldt function \Lambda(n) (which is \log p on prime powers and 0 otherwise, so \sum_n \Lambda(n) e(n\alpha) is essentially F(\alpha)) into a handful of well-structured pieces:

• "Type I" (linear) sums \sum_{d} c_d \sum_{m} e(dm\alpha), whose inner sum is a geometric series you can bound exactly;
• "Type II" (bilinear) sums \sum_{m}\sum_{n} a_m b_n\, e(mn\alpha) — the famous Vinogradov bilinear sums — where Cauchy–Schwarz turns "no cancellation known" into real cancellation.

The bilinear structure is the whole trick: even though you understand neither factor a_m nor b_n individually, the product mn ranges over enough values that Cauchy–Schwarz extracts square-root cancellation. Vaughan's identity is simply a clean, explicit way to write \Lambda as a short combination of such Type I and Type II sums. This "arithmetic-information-into-cancellation" move is one of the most reused ideas in all of analytic number theory.

Why "three" works but "two" does not

Everything comes down to how many copies of F you can spend. For three primes the minor-arc integral splits, via the sup-times-mean-square trick, as

The mean square is cheap and honest: by Parseval and the Prime Number Theorem, \int_0^1 |F(\alpha)|^2\,d\alpha \approx N\log N. That leaves just one surviving factor of F to control pointwise — and Vinogradov's estimate hands us exactly that, a saving of \sup_{\mathfrak m}|F| \ll N/(\log N)^{A}. Multiplying, the minor arcs contribute \ll N^2/(\log N)^{A-1}, which is smaller than the main term \asymp N^2/(\log N)^3 as soon as A > 4. The books balance. Three primes: proved.

Now try two primes (binary Goldbach for an even N). You want \int_0^1 F(\alpha)^2 e(-N\alpha)\,d\alpha. The same split leaves

and here the second factor is fatal. The L^1 norm \int_0^1 |F|\,d\alpha is of size roughly N/(\log N) — no honest mean-square cushion remains — so the product is about the same size as the main term itself. The error is as big as the answer. There is not enough cancellation to spare: with only two copies of F, the third power that made the minor arcs negligible is gone. This is the exact analytic reason the circle method proves the ternary problem but not the binary one.

From "sufficiently large" to every odd N ≥ 7

Vinogradov's original proof was ineffective: it guaranteed a threshold N_0 beyond which every odd number works, but — because it leaned on the Siegel–Walfisz theorem, whose constant is itself ineffective thanks to possible Siegel zeros — it could not say how large. Later workers made N_0 explicit, but astronomically so: K. Borozdkin's 1956 bound was about e^{e^{16.038}}, a number with millions of digits, hopelessly far beyond any computer check of the small cases.

The gap between "provably true past N_0" and "checkable by computer below N_0" was finally closed in 2013 by Harald Helfgott. By sharpening the major- and minor-arc estimates dramatically he pushed the analytic threshold down to about 10^{27}, then verified every odd number below that computationally. The ternary Goldbach conjecture is now a theorem for all odd N \ge 7 — unconditionally, with no "sufficiently large" left in it.