I’ve just uploaded to the arXiv my paper “Every odd number greater than 1 is the sum of at most five primes“, submitted to Mathematics of Computation. The main result of the paper is as stated in the title, and is in the spirit of (though significantly weaker than) the even Goldbach conjecture (every even natural number is the sum of at most two primes) and odd Goldbach conjecture (every odd natural number greater than 1 is the sum of at most three primes). It also improves on a result of Ramaré that every even natural number is the sum of at most six primes. This result had previously also been established by Kaniecki under the additional assumption of the Riemann hypothesis, so one can view the main result here as an unconditional version of Kaniecki’s result.

The method used is the Hardy-Littlewood circle method, which was for instance also used to prove Vinogradov’s theorem that every sufficiently large odd number is the sum of three primes. Let’s quickly recall how this argument works. It is convenient to use a proxy for the primes, such as the von Mangoldt function , which is mostly supported on the primes. To represent a large number as the sum of three primes, it suffices to obtain a good lower bound for the sum

By Fourier analysis, one can rewrite this sum as an integral

where

and . To control this integral, one then needs good bounds on for various values of . To do this, one first approximates by a rational with controlled denominator (using a tool such as the Dirichlet approximation theorem) . The analysis then broadly bifurcates into the *major arc* case when is small, and the *minor arc* case when is large. In the major arc case, the problem more or less boils down to understanding sums such as

which in turn is almost equivalent to understanding the prime number theorem in arithmetic progressions modulo . In the minor arc case, the prime number theorem is not strong enough to give good bounds (unless one is using some extremely strong hypotheses, such as the generalised Riemann hypothesis), so instead one uses a rather different method, using truncated versions of divisor sum identities such as to split into a collection of linear and bilinear sums that are more tractable to bound, typical examples of which (after using a particularly simple truncated divisor sum identity known as Vaughan’s identity) include the “Type I sum”

and the “Type II sum”

After using tools such as the triangle inequality or Cauchy-Schwarz inequality to eliminate arithmetic functions such as or , one ends up controlling plain exponential sums such as , which can be efficiently controlled in the minor arc case.

This argument works well when is extremely large, but starts running into problems for moderate sized , e.g. . The first issue is that of logarithmic losses in the minor arc estimates. A typical minor arc estimate takes the shape

when is close to for some . This only improves upon the trivial estimate from the prime number theorem when . As a consequence, it becomes necessary to obtain an accurate prime number theorem in arithmetic progressions with modulus as large as . However, with current technology, the error term in such theorems are quite poor (terms such as for some small are typical, and there is also a notorious “Siegel zero” problem), and as a consequence, the method is generally only applicable for very large . For instance, the best explicit result of Vinogradov type known currently is due to Liu and Wang, who established that all odd numbers larger than are the sum of three odd primes. (However, on the assumption of the GRH, the full odd Goldbach conjecture is known to be true; this is a result of Deshouillers, Effinger, te Riele, and Zinoviev.)

In this paper, we make a number of refinements to the general scheme, each one of which is individually rather modest and not all that novel, but which when added together turn out to be enough to resolve the five primes problem (though many more ideas would still be needed to tackle the three primes problem, and as is well known the circle method is very unlikely to be the route to make progress on the two primes problem). The first refinement, which is only available in the five primes case, is to take advantage of the numerical verification of the even Goldbach conjecture up to some large (we take , using a verification of Richstein, although there are now much larger values of – as high as – for which the conjecture has been verified). As such, instead of trying to represent an odd number as the sum of five primes, we can represent it as the sum of three odd primes and a natural number between and . This effectively brings us back to the three primes problem, but with the significant additional boost that one can essentially restrict the frequency variable to be of size . In practice, this eliminates all of the major arcs except for the principal arc around . This is a significant simplification, in particular avoiding the need to deal with the prime number theorem in arithmetic progressions (and all the attendant theory of L-functions, Siegel zeroes, etc.).

In a similar spirit, by taking advantage of the numerical verification of the Riemann hypothesis up to some height , and using the explicit formula relating the von Mangoldt function with the zeroes of the zeta function, one can safely deal with the principal major arc . For our specific application, we use the value , arising from the verification of the Riemann hypothesis of the first zeroes by van de Lune (unpublished) and Wedeniswki. (Such verifications have since been extended further, the latest being that the first zeroes lie on the line.)

To make the contribution of the major arc as efficient as possible, we borrow an idea from a paper of Bourgain, and restrict one of the three primes in the three-primes problem to a somewhat shorter range than the other two (of size instead of , where we take to be something like ), as this largely eliminates the “Archimedean” losses coming from trying to use Fourier methods to control convolutions on . In our paper, we set the scale parameter to be (basically, anything that is much larger than but much less than will work), but we found that an additional gain (which we ended up not using) could be obtained by averaging over a range of scales, say between and . This sort of averaging could be a useful trick in future work on Goldbach-type problems.

It remains to treat the contribution of the “minor arc” . To do this, one needs good and type estimates on the exponential sum . Plancherel’s theorem gives an estimate which loses a logarithmic factor, but it turns out that on this particular minor arc one can use tools from the theory of the large sieve (such as Montgomery’s uncertainty principle) to eliminate this logarithmic loss almost completely; it turns out that the most efficient way to do this is use an effective upper bound of Siebert on the number of prime pairs less than to obtain an bound that only loses a factor of (or of , once one cuts out the major arc).

For estimates, it turns out that existing effective versions of (1) (in particular, the bound given by Chen and Wang) are insufficient, due to the three logarithmic factors of in the bound. By using a smoothed out version of the sum , for some suitable cutoff function , one can save one factor of a logarithm, obtaining a bound of the form

with effective constants. One can improve the constants further by restricting all summations to odd integers (which barely affects , since was mostly supported on odd numbers anyway), which in practice reduces the effective constants by a factor of two or so. One can also make further improvements in the constants by using the very sharp *large sieve inequality* to control the “Type II” sums that arise from Vaughan’s identity, and by using integration by parts to improve the bounds on the “Type I” sums. A final gain can then be extracted by optimising the cutoff parameters appearing in Vaughan’s identity to minimise the contribution of the Type II sums (which, in practice, are the dominant term). Combining all these improvements, one ends up with bounds of the shape

when is small (say ) and

when is large (say ). (See the paper for more explicit versions of these estimates.) The point here is that the factors have been partially replaced by smaller logarithmic factors such as or . Putting together all of these improvements, one can finally obtain a satisfactory bound on the minor arc. (There are still some terms with a factor in them, but we use the effective Vinogradov theorem of Liu and Wang to upper bound by , which ends up making the remaining terms involving manageable.)

## 155 comments

Comments feed for this article

13 May, 2012 at 9:55 am

陶哲轩接近证明哥德巴赫猜想 | 博鲨科技 跨界畅游|中国领先的网站制作、电子商务、在线交易方案提供商[…] 陶哲轩最近在预印本网站arXiv.org发 表论文（已递交到《Mathematics of Computation》），证明每个大于1的奇数可以表示成最多五个质数之和，他表示有希望将数 目从五减少到三。 Leave a Reply 点击这里取消回复。 […]

13 May, 2012 at 4:19 pm

Panu HorsmalahtiMinor typo in page 36, it should say “used in the proof”.

[Thanks, this will be corrected in the next version of the ms.]13 May, 2012 at 5:09 pm

Michael Van Der KolffMinor typo on page 10, Lemma 3.1: alpha should be a non-integral real, not a coset of Z in R.

[Actually, we are identifying the two; see Section 2, especially the paragraph near (2.1).]14 May, 2012 at 5:38 am

Michael Van Der KolffYeah, I noticed when rereading it on the train heading down to class. It’s actually really quite interesting – and I hope I can understand the techniques by the end of my degree :)

It’s really quite fascinating, though :)

13 May, 2012 at 11:49 pm

Some math | pyd's blog[…] problem a relatively recent update on Goldbach conjecture has be given by a Fields medalist. https://terrytao.wordpress.com/2012/02/01/every-odd-integer-larger-than-1-is-the-sum-of-at-most-five-… Share this:TwitterFacebookLike this:LikeBe the first to like this […]

14 May, 2012 at 7:02 pm

La conjetura Golbach, ¿cerca de resolverse? | Matuk.com[…] Fuente: Terence Tao blog input, textarea{} #authorarea{border:#9EBAC7 1px solid; margin:10px 0px 30px; padding:6px 8px 14px 6px; background-repeat:no-repeat; color:#555; background-color:#e2eaee} #authorarea h3{font-size:14px; color:#333; font-weight:bold; margin:0} #authorarea h3 a{text-decoration:none; color:#333; font-weight:bold} #authorarea img{margin:0; float:left; border:1px solid #ddd; width:60px; height:60px; margin-right:12px} #authorarea p{color:#333; margin:0} #authorarea p a{color:#333} .authorinfo{ } […]

20 May, 2012 at 1:48 pm

AnonymousHi, I think the proof is wrong.

21 May, 2012 at 12:44 pm

lup@Anonymous[20 May, 2012 at 1:48 pm] as a consolation, the world you are living in where the proof is wrong must certainly have some funny sides.

20 May, 2012 at 2:58 pm

Heuristic limitations of the circle method « What’s new[…] famous theorem that every sufficiently large odd integer is the sum of three primes; my own result that all odd numbers greater than can be expressed as the sum of at most five primes is also proven by essentially the same method (modulo a number of minor refinements, and taking […]

22 May, 2012 at 8:10 pm

The Goldbach Conjecture | OU Math Club[…] Recently, Terry Tao proved that every odd integer is the sum of at most 5 primes. The key thing is that he doesn’t have to assume the validity of the Riemann Hypothesis to do it. So Kaniecki’s theorem is true without any qualifications. It’s a very nice result and Dr. Tao posted about it on his blog.** […]

23 May, 2012 at 10:33 am

Goldbach Conjecture and Terence Tao » IMatDb[…] 次日，Tao 在他的博客 之中，写了一篇日志 描述了证明的大概轮廓。 […]

23 May, 2012 at 10:36 am

Terence Tao come closer to solving Goldbach’s weak conjecture » IMatDb[…] 次日，Tao 在他的博客 之中，写了一篇日志 描述了证明的大概轮廓。 […]

29 May, 2012 at 10:28 am

El difícil ascenso a la solución de un problema matemático | El Enciclopedista[…] estadounidense había presentado en febrero de este artículo a una revista para su publicación y la experiencia, pero la revista Scientific […]

7 June, 2012 at 4:55 pm

Terence Tao come closer to solving Goldbach’s weak conjecture » Zyymat[…] 次日，Tao 在他的博客 之中，写了一篇日志 描述了证明的大概轮廓。 […]

24 June, 2012 at 9:45 am

Anon8891Hi there,

I don’t understand one thing, how does this proof work for all odd integers greater than 1 and not rely upon the Riemann hypothesis?

The number of odd integers greater than 1 is infinite.

This proof relies upon the numerical verification of the Riemann hypothesis up to some height.

Since Kaniecki proved that every odd integer is a sum of at most five primes assuming that the Riemann Hypothesis is true, how does this proof differ much from Kaniecki’s since it relies upon the numerical verification of the Riemann hypothesis up to a certain height?

24 June, 2012 at 10:20 am

Terence TaoKaniecki’s argument and mine use the Riemann hypothesis (or partial numerical verifications of it) in somewhat different ways. Kaniecki uses a quantitative version of the a well-known fact that if the Riemann hypothesis was true, then almost all prime gaps would be very small, which (together with numerical verification of the even Goldbach conjecture) implies that almost all odd numbers (in a certain range) are the sum of three primes, and hence that all even numbers (in a similar range) are the sum of six primes. (I’m glossing over a number of details here.) Unfortunately this argument doesn’t work too well with only a partial numerical verification of RH, more or less for the reason you point out, namely that one has to verify the conjecture for an infinite number of numbers. (Actually, thanks to the Vinogradov theorem results of Liu and Wang and others, one only has to verify the numbers up to 10^{1300} or so, but even in this range Kaniecki’s argument would still require far more numerical verification of the RH than is currently available.)

In my argument, numerical verification of RH is used in a slightly different way, to get a reasonably good error term in the prime number theorem, which in turn leads to control of prime exponential sums at very low frequencies. This, by itself, is nowhere near enough to settle the conjecture (in contrast to Kaniecki’s argument, in which RH is the main tool), again because the numerical verification of RH falls well short of what would be needed to cover the range of numbers desired without additional tools; the estimates from numerical RH have to be combined with effective minor arc estimates (to handle medium frequencies) and numerical verification of Goldbach (to handle large frequencies) in order to cover all the contributions to the counting function associated to this problem.

26 June, 2012 at 9:42 am

ChanakyaAccording to the preposition, every odd integer greater than 1 is the sum of “at most” five primes. But, if we had an odd number of odd primes, their sum will always be odd. So, it could be possible that an odd number be a sum of 7 odd primes or 9 odd primes.Then the preposition no more remains true. For example : 3+5+7+11+13+17+19= 75. But 75 here is a sum of 7 primes. ?????????

26 June, 2012 at 7:19 pm

Anonymousplease learn more math before you comments

18 February, 2013 at 3:40 pm

anonYou don’t understand what’s he saying.

He’s not saying that “odd integers cannot be the sum of more than 5 primes”

He’s saying that “all odd integers are the sum of 5 primes or less”

Meaning that every odd integer can be expressed as the sum of 5 primes or less.

There are multiple ways that one can gain the sum of 75 using only 5 prime numbers or less.

Like:

75 = 73+2

75 = 67+3+5

75 = 37+19+17+2

75 = 23+19+17+13+3

9 July, 2012 at 8:31 am

AnonymousVinogradov three primes theorem implies this result. This does not break any record.

13 July, 2012 at 8:26 am

AnonymousVinogrodov’s three primes theorem iplies that even integer are sums of four or less primes.

To beat the Vinogradov’s result, you will have to prove that even integers are sums of three or less primes.

13 July, 2012 at 8:57 am

Terence TaoVinogradov’s theorem is only applicable for

sufficiently largeodd natural numbers, not forallodd natural numbers; see for instance http://en.wikipedia.org/wiki/Vinogradov's_theorem .13 July, 2012 at 10:46 pm

SilvioHi Tao, do you think that larger numerical verification can be useful to improve this result? if yes, have you an idea of what exponent we must reach at least?

Thank you and congratulation for the result.

Silvio

14 July, 2012 at 11:32 am

Terence TaoSee my previous comment. Basically, by using more numerics, one can reduce the reliance on Vinogradov theorems, but it does not seem easy to get from five primes to four with this method.

6 September, 2012 at 12:46 am

ciroHi Terry, congratulation for your interesting paper. What abaout the following approach for the strong version of the conjecture?

http://arxiv.org/abs/1208.2244

Thank you

Ciro

14 September, 2012 at 5:43 am

Michael RubinsteinJust came across this. Regarding inputting numerical verification of RH- most verifications of RH have been experimental and cannot be used as the basis of proof of a rigorous mathematical statement. That includes the computations referenced.

Luckily, though, RH has been more rigorously verified, by Booker and Platt up to height 30, 610, 046, 000 (103, 800, 788, 359 zeros). http://arxiv.org/pdf/1203.5712.pdf

They used explicit truncation bounds and interval arithmetic to bound accumulated round off. In principle, their work provides a rigorous verification of RH. However, there is of course a lot that takes place in a black box (ex intel chips) or at a very low level (compilers, assembler). Hardware has never been certified rigorously, and, there have been cases of buggy hardware uncovered experimentally. Compilers also are constantly having bug fixes (look at the version history of any compiler).

There are some easy test that one can do with zeta zeros to increase one’s level of confidence. For me, testing the explicit formula with many test functions gives me confidence that a given zeta-zeros computation is correct.

Anyhow, Terry, this work sounds interesting.

14 September, 2012 at 5:55 am

Michael RubinsteinOh, I see that you do refer to Platt’s computation in your arxiv paper (I incorrectly referred, above, to the paper as Booker and Platt’s rather than Platt’s).

1 October, 2012 at 6:29 pm

QuoraWhat are the greatest open problems in math are we closest to solving?…Terence Tao recently proved that every odd integer larger than 1 is the sum of at most five primes. This advance pushes closer to the odd Goldbach conjecture, which replaces five primes with three primes. He said in a talk recently that we are probably…

23 January, 2013 at 1:51 am

The same anonymous as always xPIs the paper already being published?

Today appears in the news (SUPER SCHOLAR test) that Terrence is one of the most inteligent HUMAN BEING alive (IQ 230).

With that IQ is no merit to find such math wonders ! ( xD ) Congrats!

23 January, 2013 at 2:57 pm

????What I said I true, at least that THIS Terence Tao has appeared in the news, with such IQ. Is not worth to congratulate him -no kidding-? http://www.superscholar.org/smartest-people/

18 February, 2013 at 3:59 pm

anonTerence Tao’s adult IQ has never been tested, his 230 score was gained from childhood or teen IQ estimates (which are multiplied numbers based on ratios)

I don’t think Tao cares much about IQ either way.

25 January, 2013 at 1:05 pm

Brian McCormickDear Professor Tao,

I have seen several references saying Borozkin was the first to find an explicit value for the Vinogradov bound for the odd Goldbach conjecture and he showed the value to be 3^3^15 to be it (that is 3 to the 3rd power to the 15th power).

However, this result is not published. I would love to figure out how this bound was determined. Do you know of any reference that shows how this bound is determined? Is it so simple that you can give a brief, high-level explanation for it on here?

Congratulations on your proof. It is indeed a work of art!

Brian McCormick

29 March, 2013 at 1:05 am

freepublicDear Professor Terry Tao,

the proof has many special theorem, i can’t understand fully. i wonder what is the large even number?

if 2 is large number, it can’t express as any sum of primes.

if inifite is large number, it has 1 add number being inifite always.

you can look at my idea in http://vixra.org/abs/1301.0129

Sencerely,

Liu Ran

7 April, 2013 at 1:03 am

freepublicDear Professor Terry Tao,

you can look at my new article in http://vixra.org/abs/1304.0020

both experiment and logic have demonstrated prime being infinite is wrong.

it means Euclid’s proof had a issue which had misled machemacian for many years.

i admire you, wish you can discuss this problem. maybe it’s a big event.

Sencerely,

Liu Ran

12 April, 2013 at 1:18 am

freepublicCongratulations!

14 May, 2013 at 4:57 am

Three Primes Make an Odd | GCP Archived[…] https://terrytao.wordpress.com/2012/02/01/every-odd-integer-larger-than-1-is-the-sum-of-at-most-five-… […]

15 May, 2013 at 12:15 am

freepublici wonder why need to exclude 5 from your theorem, if it is really true.

1 June, 2013 at 3:41 am

grutgeHas been already published in a peer reviewed journal this article?

19 August, 2013 at 3:53 am

neymetmsДорогой Тао , если взять неравенство A<B<С ,и если выполнятся.

условия для A и С, то они должны выполняются и для B.

Слабая проблема Гольдбаха (любое нечётное число, превосходящее единицу, можно представить в виде суммы не более чем трёх простых.) выполняется для малых нечётных чисел. (A) не превосходящих е28. и «достаточно больших» нечётных чисел(С), то почему оно (условие) не должноне выполнятся и для (B) ?

3 September, 2013 at 10:24 am

Bringing A Riemannian Gun to a Euclidean Knife Fight |[…] Today, part of the appeal of prime numbers is that there remain many apparently-simple questions about primes whose answers are unknown. (One famous example is Goldbach’s conjecture, posed in 1742 in a letter to Leonhard Euler, that every even integer > 2 can be written as the sum of two primes. Note, however, that Terence Tao has recently proven that every odd integer > 1 is the sum of at most 5 primes). […]

30 October, 2013 at 4:56 pm

Huen Yeong KongOdd integer = 17+11+23+5+29+3+31 = 119 derived from the

following series of twin-primes anchored by 17:

f(17,17,17)+f(17,11,23)+f(17,5,29)+f(17,3,31)

This is the sum of seven primes and is odd.

The sum has more than five primes as stipulated by you.

Am I wrong?

Huen Y.K.

31 October, 2013 at 2:33 am

Anonymous119 = 5 + 43 + 71. The theorem says it is possible to write 119 as a sum of at most five primes, not that every way to write it will have at most five primes.

1 November, 2013 at 8:42 pm

Huen Yeong KongYes, you are right for my example, but could you recommend an

algorithm which could predict for every odd integer a sum of at most five primes?

Huen Y.K.

1 November, 2013 at 10:56 pm

Michael van der KolffIn general, I can’t see a way beyond generating all the primes beneath that number, and then trying each partition up to 5 of them. I’m pretty sure that’s *not* polynomial time, though.

2 November, 2013 at 5:22 pm

Huen Yeong KongYou are right. Sounds like a Four-Colour Theorem variant.

Huen Y.K.

6 November, 2013 at 7:26 am

MattForgive my stupidity – this assumes 1 is prime?

Otherwise 11 cannot be expressed as a sum of primes at all.

Are there any other prime numbers for which this does not hold if 1 is excluded?

6 November, 2013 at 11:35 am

Michael van der Kolff11 (it’s prime!), or 3+3+5. Up to 5, not exactly 5…

7 November, 2013 at 12:06 pm

MattThanks for your reply.

I thought the primes in the sum could be used only once:

eg 7 = 5 +2; 13 = 11 + 2 (not 3+3+3+2+2); 17 = 7 + 5 + 3 +2; 19 = 17 + 2 etc

It seems that (excluding 2 – the first prime), 11 is the only prime that cannot be expressed as a sum of primes (using each only once).

Wondered whether this was true.

Thanks again

7 November, 2013 at 12:54 pm

MattAnd 3!

8 November, 2013 at 2:11 am

Huen Yeong KongPlease test this Maxima line on Maxima Online:

for n from 2 thru 20 do print(sum(f(n,n-k,n+k)*(is(equal(primep(n-k)*primep(n+k),true^2))-unknown)/(true-unknown),k,0,n));

First six outputs are shown here. all are twinprimes including identical primes.

f(2,2,2)

f(3,3,3)

f(4,3,5)

f(5,5,5)+f(5,3,7)

f(6,5,7)

f(7,7,7)+f(7,3,11)

These show that every prime is half the sum of two identical copies of itself. Therefore 1 should qualify as a prime i.e. 1 = (1+1)/2.

Huen Y.K.

8 November, 2013 at 5:11 am

Michael van der KolffIt’s simply a question of what’s a useful definition. In stating the fundamental theorem of arithmetic, for example, and lots of others, we’d have to say “every prime except 1”, and when stating results about prime ideals we might have to say (when treating the ring as a ZZ module) “every prime apart from 1”, etc. So instead of doing that, we just exclude 1 from the definition.

Of course, if for your purposes you wish to define it to mean something different, then the only thing you have to do is define it that way in your paper.

8 November, 2013 at 4:00 am

Huen Yeong KongFor more information on the above, please read my published paper:

Aditi Journal of Computational Mathematics

Volume 1 ( 2013 ) , Number 1

Previous | Next | Back

Development of General Twin-Prime Number System by the Method of Indirect Induction

Huen Yeong Kong

Pages 1-12

View Details Abstract References

9 November, 2013 at 3:39 pm

MattThanks for your thoughts (and time) – personally I don’t think 1 is a prime as it can be described as n to the power of 0

For the purposes of my conjecture I would exclude 1 as a prime and say that 2,3 and 11 are the only prime numbers that cannot be expressed as a sum of a number of unique primes (ie only using each prime once).

Thanks again

9 November, 2013 at 4:23 pm

AnonymousYour remark on 2 and 3 are correct but not on 11 unless I interprete

incorrectly.. Here are the evidence::

f(11,11,11)+f(11,5,17)+f(11,3,19)

Therefore:

11=(11+11)/2, 11=(5+17)/2, 11+3+19)/2

Huen y.K.

9 November, 2013 at 7:47 pm

MattSorry I don’t understand Huen

11 cannot be expressed as a sum of lower unique primes (excluding 1)

There are only 3 ( 2,3,7)

eg

2 + 3 + 5 + 7 = 17

2 + 3 + 7 = 12

2 + 7 = 9

2 + 5 = 7

etc

9 November, 2013 at 7:52 pm

Matt13 on the other hand can be expressed as:

2 + 11

or 7 is:

2 + 5

19 is:

2 + 17 etc

9 November, 2013 at 9:02 pm

AnonymousYes, your are right. I interpret wrong. I do agree that 1 should not be

defined as a prime because if it is done all primes from 2 to infinity

will become composites, e.g.:

factor((3) = 1*3

factor(23)=1*23 …

The whole natural number sequence except 1 will be composite only.

HuenYK

7 December, 2013 at 1:01 pm

JasonWhat am I missing.

3 is odd and cannot be represented by the summation of primes.

11 is odd and cannot be represented by the summation of primes.

Doesn’t seem to be accurate.

9 December, 2013 at 12:42 pm

MattThat was my point 2,3 & 11 are the only primes that cannot be expressed as a sum of lower unique (ie used only once in the sum) prime numbers.

17 December, 2015 at 7:54 am

Anonymous633 = 139+211+283 3 primes

633 = 131+5+3+211+283 5 primes

633 = 131+5+3+181+17+13+283 7 primes

633 = 131+5+3+181+17+13+163+61+59 9 primes

633 = 127+2+2+5+3+181+17+13+163+61+59 11 primes

633 = 53+41+37+5+3+181+17+13+163+61+59 11 primes

633 = 83+19+29+5+3+181+17+13+163+61+59 11 primes

633 = 61+47+23+5+3+181+17+13+163+61+59 11 primes

and 11 primes is not the upper limit for 633. In fact a necessary and sufficient condition for an odd number N to be sum of 2k+1 primes is N greater than 4k+3.. Hence the numbers 3, 5, 7 and 9 can not be sum of 5 primes. the first odd number to be sum of 5 primes is 11, it happens it is a prime!