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Can any even number be written in the form of $(n-1)A+B$, where $A$ and $B$ are primes and $n$ is an even number? $$ C= (n-1)A+B $$ Here $C$ is a even number and $A$ and $B$ are prime and $n$ is any even number such that $n < \sqrt{C}$ (or you can say that $n$ is the number of primes used in the eqn)

Is this true? Because if Goldbach's conjecture is true, I presume this should also be true, right?

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what about C=6, n=4? –  wxu Dec 8 '12 at 16:38
    
@wxu: I think the OP is asking if this can be done for every even number, not just for some (positive) even number. –  hardmath Dec 8 '12 at 16:43
    
@wxu May be I can modify that stating n should be less than the squareroot of c [n < sqrt(c)] –  Shan Dec 8 '12 at 17:02
    
What about C=66, n=4 for your modifying case? –  wxu Dec 8 '12 at 18:46
    
@Wxu n can be any even number than is less than 8 so it can be also 6 and need not be 4 and it can be also 2 so that 66=(6-1)11+11 –  Shan Dec 9 '12 at 4:59
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1 Answer

up vote 2 down vote accepted

Chen's theorem, that every sufficiently large even number is the sum of a prime and another number with at most two prime factors, is a result in this direction. That is, taking $n=2$ in your expression accounts for those even numbers which are a sum of two (odd) primes, and taking $n=p+1$ where an even number is a sum of a prime and a product $pq$ of two odd primes (semiprime or "almost prime") accounts for the others.

Some work has been done to improve on Chen's work, but I don't know of an explicit bound beyond which his sieve methods show this representation holds.

Edit: The construction above almost meets the bound $n^2 < C = 2m$ if we choose $p$ to be the lesser prime factor of the semiprime summand (if applicable). Unless the factors are equal (the semiprime is a square), $(p+1)^2 < C$.

Without the restriction on $n$ I don't think your result is difficult to prove for $C \gt 4$. Consider the so-called Bertrand's postulate that a prime exists between $m$ and even number $2m$. Let $q$ be the least such prime, so that $2m-q$ is an odd number. If this is prime, we are done (as explained above for $n=2$ in your notation). If this is an odd number greater than $3$, then is has an odd prime factor $p$, and again we are done as before ($2m = ((2m-q)/p)*p + q$ and the factor $(2m-q)/p$ is (being odd) one less than an even $n$).

The only difficulty I can see is with very small even numbers. That is, we cannot express $2$ or $4$ as sums of two (positive) odd primes, nor can we express them as a sum of the kind you ask about (if we restrict ourselves to positive summands).

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I see that some restrictions on the size of $n$ have been added to the Question since I first posted, so my "answer" may not apply to the revised version. Certainly restricting $n$ makes the Question more interesting. –  hardmath Dec 8 '12 at 17:17
    
Wxu's comment made me think more about it –  Shan Dec 8 '12 at 17:53
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