# Prove by contradiction that every integer greater than 11 is a sum of two composite numbers

I have thought a lot but am failing to arrive at anything encouraging.

First try: If this is to be proved by contradiction, then I start with the assumption that let $n$ be a number which is a sum of two numbers, of which at least one is prime. This gives $n = p + c$, where $p$ is the prime number and $c$ is the composite number. Also, any composite number can be written as a product of primes. So I can say, $n = p + p_1^{e_1}.p_2^{e_2}...p_k^{e_k}$. From this, I get $n - p = p_1^{e_1}.p_2^{e_2}...p_k^{e_k}$, but I have no clue what to do next.

Second try: For an instant let me forget about contradiction. Since $n > 11$, I can say that $n \geq 12$. This means that either $p \geq 6$ or $c \geq 6$. Again I'm not sure what to do next.

Finally, consider that the number 20 can be expressed in three different ways: $17+3$ (both prime), $16+4$ (both composite), and $18+2$ (one prime and one composite). This makes me wonder what we are trying to prove.

The textbook contains a hint, "Can all three of $n-4$, $n-6$, $n-8$ be prime?", but I'm sure what's so special about $4, 6, 8$ here.

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At least one of the three numbers $n-4$, $n-6$, $n-8$ is divisible by a certain prime... –  anon Jul 24 '13 at 9:08
(what we are trying to prove is that it exists at least one way to write a number greater than 11 as the sum of two composite numbers. You may partition it in many different ways: what matters is, at least one partition uses two composite numbers) –  mau Jul 24 '13 at 10:15

Spoiler #1

You can write $n = (n - \varepsilon) + \varepsilon$, where $\varepsilon \in \{4, 6, 8\}$.

Spoiler #2

$n - \varepsilon > 3$, as $n > 11$.

Spoiler #3

One of the three numbers $n - \varepsilon$ is divisible by $3$, as they are distinct modulo $3$.

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Spoiler #4 >! nice spoiler(+1) –  Sami Ben Romdhane Jul 24 '13 at 9:07
@SamiBenRomdhane, thanks! –  Andreas Caranti Jul 24 '13 at 9:07
This is great! But where does this involve proof by contradiction? –  dotslash Jul 24 '13 at 9:26
It doesn't. So what? Why do you care how it's proved? –  Gerry Myerson Jul 24 '13 at 9:36
@Gerry Presumably because that's what's written in the textbook OP mentions (why it would say it wants a proof by contradiction given the nature of its hint I don't understand). –  anon Jul 24 '13 at 10:51