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I have seen the following problem:

Any number $n>11$ can be formed by the sum of two composite numbers

and it says to prove it by contradiction. I have done the following:

Assume that the sum will be formed by non-composite numbers. Because a composite number is one that has at least one divisor that is not one or the same number; I can assume that it could be the sum of two prime numbers (because these are not composite). From this point I can say that adding both numbers which are primes, a and b, will give us necessarily an odd number. So: $a=2p+1$ and $b=2p+1$, adding both I will have:

$a+b=2\cdot 2p + 2 = 2(2p+2)$ which is an even number, leaving any odd number; and giving us a contradiction that our theorem will hold for any n>11.

A similar proof can be done if we suppose that one number is composite and the other not, in this case any even number will be left out.

is this fine?

Thanks

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1 Answer

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No, it is not fine. If $n$ is not the sum of two composite numbers, it does not follow that it must be the sum of two primes. It might be the sum of a prime and a composite number. All you can say is that it cannot be expressed as the sum of two composite numbers. Also 1 is not regarded as either prime or composite.

Also, you are not asked to show that all numbers from some point on are the sum of two composite numbers, but specifically to show that all numbers greater than 11 are the sum of two composite numbers.

You are asked to prove it by contradiction. But you probably also need to use induction. It is hard to think of a way without induction. So the kind of thing you want to prove is that at least one of $n-4,n-6,n-8$ must be composite.

However, you did not ask for a proof, just for a critique of what you had done. So I will leave you with that hint.

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