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

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 $3$.

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!!

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.