# Is every sufficiently large even integer the sum of distinct primes?

Is every sufficiently large even integer the sum of (any number of) distinct primes?

No doubt this question has been asked before; does the conjecture/theorem have a name? It is related to Goldbach's conjecture, which states that every even integer is the sum of two (not necessarily distinct) primes.

The even integer $6$ can not be expressed as a sum of distinct primes as its only prime representations are $2+2+2$ and $3+3$, thus the condition "sufficiently large" is necessary. If the condition "distinct" is removed, the statement becomes obvious ($n$ copies of $2$).

• If Goldbach's conjecture is true, then either $n/2$ is even, so $n=p+q$ and $p\neq q$, or else $n=2+p+q$ and $p\neq q$ – Empy2 Aug 27 '15 at 13:55

The contraint to have a representation as a sum of distinct primes does not really affect the magnitude of $r_k(n)$, i.e. the number of ways to write $n$ as a sum of $k$ primes.
Vinogradov's theorem implies that every sufficiently big odd number is the sum of three primes (since it gives a lower-bound for $r_3(n)$), hence every sufficiently big even number is the sum of four distinct primes.
• Adopting the convention of 1 as a prime, you can solve some things, in particular 6 = 2 + 3 + 1. On the other hand, Golbach can be stated the following suggestive way: all natural integer greater than 1 is equidistant from two primes (making few exceptions with small numbers, for example 3). This simply because $2n=p+q$ is equivalent to $n-p=q-n$. – Piquito Aug 27 '15 at 15:36