# Number Theory

1. Sep 12, 2008

### tanzl

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.

2. Sep 12, 2008

### entorm

I don't see how you used the fact that m is greater than 11. It does not appear in your calculation.

I have not done a complete calculation, but I assume you could use the property:
'all prime numbers above q are of form q#·n + m, where 0 < m < q, and m has no prime factor ≤ q' (q# means the product of the primes up to q)