For every natural integer $N>3$ there are at least two distinct prime numbers $p$ and $q$ such that $\dfrac{p+q}{2}=N$ and $N-p=q-N$, $(p<q)$.

Question

I'm not sure but this problem may be similar or related to Goldbach conjecture? Any proof/disproof, insight and opinion is appreciated, thanks.

Answer 1

First note that the two equations you give are equivalent; $$N-p=q-N\quad\Leftrightarrow\quad 2N=p+q\quad\Leftrightarrow\quad N=\frac{p+q}{2}.$$ The problem you give is slightly stronger than the Goldbach conjecture:

Suppose for every natural number $N>3$ there are two distinct prime numbers $p$ and $q$ such that $$N=\frac{p+q}{2},$$ For every even natural number $M$ there exists a natural number $N$ such that $M=2N$. By assumption there exist distinct primes $p$ and $q$ such that $N=\tfrac{p+q}{2}$, from which it follows that $$M=2N=2\cdot\frac{p+q}{2}=p+q.$$ So every even natural number is then the sum of two distinct primes, which is slightly stronger than Goldbach's conjecture.