# 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.

1. Is it true? I tried proving the nontrivial direction but got stuck.
2. 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).

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are you sure that the inequality 0 < m (< n) is strict? Let p = q = 2. then for n = 2, we have that $2\cdot 2 = 2 + 2$, and so $pq = 2 \cdot 2 = 4 = n^2 - m^2 = 2^2 - m^2 = 4 - m^2$, so $m$ would need to be zero, else we have $n, p, q$ for which the antecedent is try, but the consequent false. Perhaps we need $0 \leq m < n$ with $m = 0 = n - p = 2 - 2, p = q$. However, I haven't even checked to verify that the "someone's" observation is true. Also, I suspect that we are to take m, n, p, q to non-negative integers? –  amWhy Jun 29 '11 at 20:05
I suspect that he overlooked the fact that one can have $n^2-m^2=pq$ for primes $p<q$ without having $p=n-m$ and $q=n+m$, since it may be that $n-m=1$, $n+m =pq$, and $2n=pq+1$. –  Brian M. Scott Jun 29 '11 at 20:19

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.

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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.