Goldbach's conjecture and difference of squares Someone came to me with the following observation: If $2n=p+q$ then $pq=n^2-m^2$ for some value of $0<m<n$ (namely, $m=n-p$ given $p\le q$).
Now he claims that this is actually equivalent: that the claim "For every $n$ there exists $0<m<n$ such that $n^2-m^2$ is the product of two primes" is equivalent to Goldbach's conjecture.


*

*Is it true? I tried proving the nontrivial direction but got stuck.

*Is it well known? I tried looking for references and couldn't find any.


(I am trying to explain to him that this is a hard conjecture and trivial observations are probably not worth his time except for recreation).
 A: If $p$ and $q$ are primes and $m$ and $n$ are positive integers with $pq=n^2-m^2$
Then $pq = (n+m)(n-m)$
$p$ and $q$ are prime, so 
either $n+m = pq$ and $n-m=1$, which implies $2n = pq+1$
or $n+m=p$ and $n-m=q$, which implies $2n=p+q$
The question as stated does not exclude the first possibility, so the equivalence is not proven.
Note that, in the forward direction, $n-m=p$, and $p>1$.
So for $p$ and $q$ different the equivalence would work for $0<m<n-1$. However, the possibility that $p$ and $q$ are the same is then missed, so we would need to allow $m=0$ too.
So if we are given $n$ and we can find an $m$ to satisfy the revised condition, we have found two odd primes which sum to 2n.
A: If you are able to read French, you might be interested in looking at the following link, in which I try to prove that the smallest $r$ such that $n-r$ and $n+r$ both are prime is such that $r=O(\log^2 n)$. If not, I'll try to give an English translation this weekend, tonight I feel too tired to do so. Meanwhile, anyone is welcome to give the desired translation if needed.
Here comes the link: http://www.les-mathematiques.net/phorum/read.php?5,728922
