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.

Solution 1:

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 $3$.

Solution 2:

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!!

Solution 3:

Let's say that integer $n>11$ can't be expressed as the sum of two composite numbers. Then:

• $n=a+p$ (p is a prime and a is a composite or prime number)

Even numbers that greater than $2$ are composite.

The number of even numbers that smaller or equal to $n$ is $[\frac{n-2}{2}]$(Why?).

We said that $n$ can't be expressed as sum two composite numbers, then there have to be $[\frac{n-2}{2}]$ prime numbers at least(Why?).

But this result can't hold for $n\geq 30$, a contradiction.