MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This ought to be a simple one to answer. Does anyone know of, or can anyone provide, an accurate English translation of the marginal remarks in Goldbach's letter to Euler

in which a statement equivalent to the Goldbach conjecture is first stated?

share|cite|improve this question
Sorry, can't answer, but the letter seems very interesting, not least because of the mixture of German and Latin. Not being a native English speaker, I'm aware of a similar phenomenon today: I often have conversations, nominally in Spanish, which are liberally sprinkled with many English words, due to their mathematical content. Although perhaps hardly surprising, I hadn't realised that this was the case also earlier with Latin. – José Figueroa-O'Farrill May 3 '10 at 16:10
May I humbly ask: is that an indirect hint to watch for math.NT this week? – Thomas Sauvaget May 3 '10 at 18:40
Ha! I wish. No, I'm preparing for a public lecture at the Radcliffe Institute. – Ben Green May 3 '10 at 18:48
up vote 13 down vote accepted

I have an English translation of the letter; you can find my email address on my homepage. Here is the relevant part:

"By the way, I take it to be not useless to note down also such propositions that are very probable, even if a real proof is lacking; for even if afterwards they were found to be erroneous, they could all the same give occasion for the discovery of a new truth. Thus Fermat's idea that all the numbers $2^{2^{n-1}}+1$ yield a series of prime numbers cannot hold up, as you already demonstrated, Sir; but it should still be remarkable if this series were composed only of numbers that could be split into two squares in a unique way. I should like to risk another conjecture of that kind: any number composed from two primes is the sum of as many prime numbers (including $1$) as one wishes, right down to the sum that consists just of ones.

After reading this through again, I see that the conjecture can be proved quite rigorously for the case $n+1$ if it holds for the case $n$ and $n+1$ can be split into two prime numbers. The proof is very easy; and at least it appears to be true that every number greater than $2$ is the sum of three prime numbers."

The translation was done by Martin Mattmüller.

share|cite|improve this answer
Franz: that might be helpful? – Ben Green May 3 '10 at 19:27
Done . – Franz Lemmermeyer May 4 '10 at 5:41
Brilliant. And we could all learn from his style... – Ben Green May 4 '10 at 12:34

One of the problems seems to be the handwriting - at least here is a copy which is transcribed, perhaps this helps to get the gist (the German/Latin is even for today's native speakers hard to understand):

share|cite|improve this answer
Brilliant - now we just need someone who reads German to take a look at the footnote at the bottom of page 127. – Ben Green May 3 '10 at 17:35
I read German but not Latin. Here's a quickly-attempted translation, for whose accuracy I cannot vouch: "After reading through this again, I find that the conjecture may be completely proved in the case n+1, if it holds for n and n+1 is the sum of two prime numbers. The proof is very easy. It appears at least that every number greater than 1 is the sum of three prime numbers." – Mark Meckes May 3 '10 at 18:00
What is "the conjecture"? Something like every number $n$ being the sum of exactly $m$ "primes" for each $3 \leq m \leq n$, counting $1$ as prime? That's my guess just looking at the formulae in his letter. In the case $n = 2k + 2$ even and $m = 3$ it implies the usual Goldbach conjecture for $2k$, unless $2k - 1$ is prime. – Ben Green May 3 '10 at 18:17
@Mark: I agree with your translation; although it's been a while since I learnt German. – José Figueroa-O'Farrill May 3 '10 at 18:17
@Ben: "that every number which is composed of two prime numbers, is an aggregate of as many prime numbers as one wishes, up to the sum of only unities", or something like that. Goldbach certainly does use a lot of different words to refer to sums. – Mark Meckes May 3 '10 at 18:27

I do not have the reputation to comment on the answer, so I have to start a new one.

Freely translating page 127 of :

I deem it to be advantageous to note the following conjecture, even though it lacks a proof, as a counterexample could provide further insights. Fermat's idea that every number of the form $2^{2^{n-1}}+1$ is prime can, as you have shown, not be true; but it would be strange if this series yielded a lot of "numeros unico modo in in duo quadratis divisibiles" (numbers that can be divided into two squares???). I, too, would like to hazard a conjecture: that every number that is the sum of two primes is the sum of arbitrary numbers of primes (or 1), except the "congierem omnium unitatum" (the collections of all in one???; the footnote was already translated by Mark), for example...

share|cite|improve this answer
It's a long time since I learnt Latin... but I think that "numeros unice modo in duo quadratis divisibiles" is "number that can be divided into [written as a sum of?] two squares in a unique way". – Tom Smith May 3 '10 at 20:49

I'm not really a language expert, but I think the last sentence of the footnote reads: "At least it seems that every number greater than 1 is an 'aggregatum trium numerorum primorum'" and 'aggregatum trium numerorum primorum' should mean "sum of three primes" (and it should really mean "sum of three or less primes")

Edit: correction, thanks to José

share|cite|improve this answer
I would have thought "wenigstens" means "at least". At least that's the modern meaning. – José Figueroa-O'Farrill May 3 '10 at 18:17
You're absolutely right, José. – Georges Elencwajg May 3 '10 at 18:30

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.