January 31st, 2016, 02:21 AM  #1 
Senior Member Joined: Nov 2010 From: Berkeley, CA Posts: 174 Thanks: 35 Math Focus: Elementary Number Theory, Algebraic NT, Analytic NT  Sum of Composites
Here's are a fairly simple problem. I know the solution and I thought others might enjoying solving it: Prove that every integer n > 11 is the sum of two composite numbers. 
January 31st, 2016, 04:48 AM  #2 
Senior Member Joined: Jul 2014 From: भारत Posts: 1,178 Thanks: 230 
Proof (by contradiction): Assume there exists a number n such that n > 11 and n is not the sum of 2 composite numbers. We have n = (n − 4) + 4. Notice that 4 = 2 · 2 is a composite number. Since, by assumption n is not the sum of two composite numbers, we conclude that (n − 4) is not a composite number. Similarly, we have n = (n − 6) + 6. Notice that 6 = 2.3 is a composite number. Since, by assumption n is not the sum of two composite numbers, we conclude that (n − 6) is not a composite number. Once more, we have n = (n − 8) + 8. Notice that 8 = 4.2 is a composite number. Since, by assumption n is not the sum of two composite numbers, we conclude that (n − 8) is not a composite number. We have proved that none of (n − 4),(n − 6) or (n − 8) is composite. By the division theorem, there exist integers q and r such that n = 3.q + r and 0 ≤ r < 3. We consider 3 cases: 1. If r = 0, then n − 6 = 3q − 6 = 3 · (q − 2) 2. If r = 1, then n − 4=3q + 1 − 4 = 3 · (q − 1) 3. If r = 2, then n − 8 = 3q + 2 − 8 = 3 · (q − 2) In all three cases, we have shown that one of (n − 4),(n − 6) or (n − 8) is a multiple of 3. Since n > 11, we also know that (n−4),(n−6) and (n−8) are all bigger than 3. It follows that one of (n−4),(n−6),(n−8) composite. But this a contradiction, because we had proved that none of (n−4),(n−6),(n−8) is a composite. So, our initial assumption that there exists an integer n > 11 which is not the sum of two composite numbers must be false. This proves that every integer bigger than 11 is the sum of two composite numbers. 

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composites, sum 
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