finding any pair $(a,b)$ which adds up to up to a composite $n$ , $n\geq20$ and $a$ and $b$ ,both are composite How do we represent a composite number $n$ where $n\geq20$ as a sum of two another composite numbers .What is the best method to find any such one  pair $(a,b)$ out of many possible pairs ?
 A: One of the numbers $n-4,n-6,n-8$ must be divisible by $3$ and hence is composite because $n-8>3$
A: The trivial solution would be: Choose $a = 0$ and $b = n$. So let's assume that $a,b \geq 1$. There is a nice way to construct $a$ and $b$ out of the factorisation of $n$:
Since $n$ is composite, we can write $n = x \cdot y$ with $1 < x,y < n$. W.l.o.g we may assume that $y \geq x$. This implies $y \geq 4$, because otherwise we had $x \cdot y \leq 3 \cdot 3 = 9 <20$. As a consequence we can find natural numbers $z_1,z_2 \geq 2$ such that $y = z_1 + z_2$. Now we get $$n = y \cdot x = (z_1 + z_2) \cdot x = z_1 \cdot x + z_2 \cdot x.$$
Since $z_1,z_2 \geq 2$, we know that $z_1 \cdot x$ and $z_2 \cdot x$ are both composite, so we can choose $a = z_1 \cdot x$ and $b = z_2 \cdot x$.
An example to illustrate the construction: Let's take $n = 27 = 9 \cdot 3$. Now we can write $9 = 2 + 7$ and get $27 = 9 \cdot 3 = (2+7) \cdot 3 = 2 \cdot 3 + 7 \cdot 3 = 6 + 21$, and both 6 and 21 are composite. Note that it is important that both summands are larger than 1. If we had $n = 9 = 3 \cdot 3$ and $3 = 1+2$, we would get $9 = 3 \cdot 3 = (1+2) \cdot 3 = 3 + 6$, but 3 is prime. But in the case $n \geq 20$ there will always be one factor which is $\geq 4$, as proved above.
A: Take the composite numbers $a=2 \cdot 5=10, b=5 \cdot 25$. Their sum is $35 \ge 20$, and their sum is composite because $10+25=5(2+5)$
A: So you want to prove any composite equal or over $20$ can be written as a sum of two composites?
Let $n = k*m\ge 20; k \le m$.  Then $m \ge \sqrt{n} > 4$.  So $m = a + b; a>1, b>1$ and $n = ak + bk$.
Do you require that the three be relative prime?  I guess not as $21$ can not be so written.
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Actually it is sufficient that $n > 9$.  $n = k*m = 2k + (m-2)k$.  The only necessity is that $m \ge 4$.  So only composite numbers that will not have this property are those with no factor larger than $3$.  $9 = 3*3$ is the largest such number.  So all composite numbers $n=k*m > 9$ can be so written. 
