1. The problem statement, all variables and given/known data Use proof by contradiction to show that every integer greater than 11 is a sum of two composite numbers. 3. The attempt at a solution The question sounds a bit awkward for me since for the numbers I have tested so far. The number can be consists of 1 prime 1 composite or both primes or both composites. So, it is a headache to prove it by contradiction since every case can be true. I have done something but I am not sure whether it can even considered to be a proof o not. Assume that m is an integer larger than 11 and consists of 1 prime and 1 composite. So, m = P + C p = m - C If m is an odd integer, 2k+1 I can always find an odd C (2l+1) such that p is a composite number. p = 2k+1-2l-1 =2(k-l) Note that I can choose C such that (k-l)>1 So, p is not a prime number. Similiarly, if m is an even number (2k). I can always find an even C (2l) such that p is a composite number. p = 2k-2l =2(k-l) Note that I can choose C such that (k-l)>1 So, p is not a prime number. Hence, contradiction for both cases. The conclusion is for integer>11. I can always find 2 composite number such that their sum is equals to the integer. But the problem is I think I can also do the same thing such that when I find an appropriate C, p is a prime number. And also for the case both are primes.