i observed that every composite no. with exception of 4 and 6 can be expressed as sum of distinct prime no.s. eg: 200=103+97 100=53+47 25=13+7+5 is this true? is there any theorem stating as such?
Interesting question! It seems true. Maybe you could prove it using induction? The base cases could be integers from 1 to 20, say, by just showing the sums. Then you could assume that it works for all composite numbers up to a composite number k. Then does it work for the next composite number? Well, look at the largest prime number less than it. If composite - prime is not 1 or 4 or 6, then the sum can certainly be expressed as prime + (sum for the difference composite - prime, which can definitely be done... if the difference is prime, just take the prime, otherwise, the inductive hypothesis covers it). If the difference is 1 or 4 or 6, try the next prime number below the one you just tried. If this isn't 4 or 6, do the same thing. If it is, try the next lower prime number. And etc. 21 = composite, 19 = prime, diff = 2, done with sum = 19 + 2. 22 = composite, 19 = prime, diff = 3, done with sum = 19 + 3. 24 = composite, 23 = prime, diff = 1. Try 19 = prime, then diff = 5 and done with 19 + 5. 25 = composite, 23 = prime, diff = 2, done with 23 + 2. 26 = composite, 23 = prime, diff = 3, done with 23 + 3. 27 = composite, 23 = prime, diff = 4, try 19 = prime, diff = 9, done with 19 + psum(9) = 19 + 7 + 2. etc. This is just a sketch, but I think it should work. Does anybody have any comments?
Yes, indeed. It also follows from Bertrand's postulate that the number of such distinct primes does not exceed log_{2}(N), where N is the original number to be decomposed. Am I correct?
This is almost identical to 'Goldbach's conjecture' which is more than two and a half centuries old, famous, and has never been proved, so you are unlikely to.
This one is 'easier' in the sense that there is no limit on the number of primes that make the sum, just that they are distinct, whilw Goldback's conjecture restricts to two primes. It seems to me that Bertrand's postulate is sufficient for the proof.