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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
In your first try, you should say that $n$ is a number such that for every way of expressing it as a sum, at least one number is prime. For example, $12$ satisfies what you say, because $12=9+3$ and $3$ is prime. You then cannot assume the sum includes a composite-both numbers can be prime. Neither of these observations go to the heart of the problem. – Ross Millikan Feb 10 '15 at 16:48
up vote 10 down vote accepted

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) – user63181 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

How about this solution??

If $n$ is even, then $n$ is of the form $2k$ where $k \geq 6$. Hence $n = 2(k-4) +8$.

And if $n$ is odd, then $n$ is of the form $2k+1$ where $k\geq5$. hence $n = 2(k -4) +9$.

Thus any number $> 11$ can be expressed as the sum of two composite numbers!!

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Only 9 even numbers greater than 4 can't be expressed as the ORDERED sum of two ODD composites, namely 6, 8, 10, 12, 14, 16, 22, 32, 38.

Look at the 4 identities: 1. pp(2n)=pr[2,n]-pc(2n) 2. cc(2n)=c[2,n]-cp(2n) 3. pp(2n)=pr[n,2n-2]-cp(2n) 4. cc(2n)=c[n,2n-2]-pc(2n)

where pp(2n)=number of ordered sum of 2 primes = 2n, cc(2n)=# of ordered sums of 2 composites=2n, cp(2n)=number of ordered sums of 1 composite and 1 prime (in that order)=2n, and pc(2n)= number of ordered sums of 1 prime and 1 composite (in that order)=2n, and a+b is an ordered sum iff a< or = to b, pr[a,b] = number of primes in[a,b], c[a,b] = number of composites in [a,b]

Lots of other identities to construct from the 4 above - have fun playing with.

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and: pr[a,b] = the number of primes in [a,b] and c[a,b]= the number of composites in [a,b] – d williams Dec 10 '14 at 23:48
For some basic information about writing math at this site see e.g. here, here, here and here. – Chantry Cargill Dec 10 '14 at 23:49

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