# Sum of primes

1. Apr 20, 2005

### metrictensor

Is it possible to express all natural numbers greater than 2 as the sum of N unique prime numbers? For example, 6 = 2 + 3 and 18 = 13 + 5.

2. Apr 20, 2005

### matt grime

6=2+3 does it?

What about the smallest natural greater than 2 that isn't a prime? isn't that a counter example too?

Last edited: Apr 20, 2005
3. Apr 20, 2005

### metrictensor

I forgot to mention that 1 and 2 can be used in the sum. I made a mistake. Not 6 = 2 + 3, 5 = 2 + 3. Once you define 1 and 2 you get 3. From 3+1 you get 4 and so on. 5=2+3....

4. Apr 20, 2005

### HallsofIvy

So you are NOT talking about a "sum of primes" because 1 is not prime.

5. Apr 20, 2005

### matt grime

Now I'm a little perplexed. 2 is a prime. Anyway, the result follows quite easily from Russell's postulate:

For every n>1 there is a prime number p such that n<p<2n.

So, given m an integer, if it is even pick an prime in the range m/2 to m, and if it is odd pick a prime in the range (m+1)/2 to m+1, call this prime p(1).

Then m-p(1) < m/2, and we proceed by indcuction - the next prime we pick must be distinct from p(1) since it must be less than m/2, and p(1) is greater than m/2.

We just need to show how to write the sum for "small m" to get the full proof, ie m=1,2,3 which yo'uve done.

6. Apr 23, 2005

### robert Ihnot

Matt grime: the result follows quite easily from Russell's postulate.

Yeah? As re-expressed by Euler, an equivalent form of this conjecture (called the "strong" or "binary" Goldbach conjecture) asserts that all positive even integers can be expressed as the sum of two primes.http://mathworld.wolfram.com/GoldbachConjecture.html

So you are saying the Goldbach Conjecture has been proven?

7. Apr 23, 2005

### Icebreaker

What if we added up the first 300 002 primes?

8. Apr 23, 2005

### Zurtex

What is that supposed to mean :uhh: ?

9. Apr 23, 2005

### Icebreaker

Nevermind.

10. Apr 23, 2005

### matt grime

Of course not, but the OP didn't ask for the sum of TWO primes, he asked for the sum to be expressed as *some* sum of N distinct primes, allowing for 1 to be included.

The possibly dodgy assumption I made was that it wasn't supposed to be a fixed N, which seems reasonable, given the other hidden assumptions in the post, after all it would be required that N=1 or 2 otherwise, which we know to be false and bloody hard respectively.

Last edited: Apr 23, 2005
11. Apr 23, 2005

### robert Ihnot

Last edited: Apr 23, 2005
12. Apr 24, 2005

### matt grime

Well, whatever it should be called, it states given any natural n greater than 2 there is a prime p satisfying n<p<2n, or similar.

13. Apr 24, 2005

### Zurtex

It has been proved there exists a prime for any natural number n > 2 there exists a prime (and I'm really dragging this one from memory) p such that:

$$n - n^{\frac{23}{42}} < p < n$$

Which is quite a bit stronger . Although I'm not sure how useful.

14. Apr 24, 2005

### CRGreathouse

Here's the formula on the Wolfram site

http://functions.wolfram.com/NumberTheoryFunctions/Prime/29/0003/