question related to Goldbach Conjecture I was wondering if there is a proof to the following statement. 
Every even number $n \gt 30$ can be expressed as a sum of a odd prime and a relatively prime odd composite number (being relatively prime to $n$). 
Examples: 26 = 15 + 11, 48 = 23 + 25; 50 = 23 + 27 etc. 
Thanks!
 A: For even $n$, let $p$ be a prime occuring in the interval $[\frac{n}{2},n-2]$, and let $q=n-p$. Then $n=p+q$, and since $q<p$ and $p$ is prime, the summands will be relatively prime odd integers. The existance of at least one $p$ is guaranteed for $n>6$ by the Strong Bertrand's Postulate. This construction may fail to give a composite $q$ for some $p$ in the interval.
If, for every prime $p$, we still obtain a prime $q$, then we have made as many partitions of $n$ into two primes as there are primes in $[\frac{n}{2},n-2]$. Let the number of these Goldbach Partitions be given by $g(n)$. This would imply then that $g(n) = \pi(n-2) - \pi(\frac{n}{2}-\frac{1}{2})$. A result on the upper bound of Goldbach Partitions (Theorem $(1)$ in the paper), shows that the last $n$ for which $g(n) = \pi(n-2) - \pi(\frac{n}{2}-\frac{1}{2})$ is $210$, and for all following $n$, $g(n)$ is less than this number. Therefore, there must exist at least one composite $q$ for all $n>210$.
For $n\le210$, it may be computationally checked to see if they have such a partition, the prime would have to occur, in these cases, in $3\le p < \frac{n}{2}$.
