Proving Every Integer > 11 is Sum of Two Composite Numbers

[tanzl]

Homework Statement

Use proof by contradiction to show that every integer greater than 11 is a sum of two composite numbers.

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.

 

[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)

 

[Architeuthis Dux]

So, I am not sure whether this is a valid proof or not.

I appreciate your attempt at a solution and your critical thinking about the problem. However, your proof is not complete and can be improved. Let me provide you with some feedback and suggestions:

1. Your proof is not a proof by contradiction. In a proof by contradiction, you assume the opposite of what you want to prove and then show that it leads to a contradiction. In your proof, you assume that m is an integer larger than 11 and consists of 1 prime and 1 composite, and then show that p is not a prime number. This is not a contradiction, as your assumption is still valid. To prove by contradiction, you need to assume that m is not a sum of two composite numbers and then show that it leads to a contradiction.

2. Your proof only considers the case where m is a sum of one prime and one composite number. As you mentioned, m can also be a sum of two primes or two composite numbers. To prove the statement for all integers greater than 11, you need to consider all possible cases.

3. Your proof only considers odd and even numbers, but m can also be a multiple of 3, 5, 7, or any other prime number. You need to show that for any integer greater than 11, there exist two composite numbers that can be added to form that integer.

4. You need to show that the two numbers you choose are actually composite. In your proof, you assume that p is a composite number, but you don't show why it is composite. To complete your proof, you need to show that p and C are both composite numbers.

5. Finally, your proof only considers one specific example (m = P + C), but to prove the statement for all integers greater than 11, you need to show that it is true for all possible integers. This means that you need to generalize your proof and show that it works for any integer greater than 11.

In summary, your attempt at a solution is a good start, but it needs to be developed further to be a complete proof. I suggest you think about how to prove by contradiction and how to consider all possible cases for m (one prime and one composite, two primes, and two composites). Also, think about how to show that the two numbers you choose are actually composite. Good luck!