Each of p and q is a prime number that satisfy this equation:
p q+1 2n
--- + ----- = -----
p+1 q n+2
where n is a positive integer.
Determine all possible value(s) of q-p.
Starting with some algebraic manipulation:
p/(p+1) + (q+1)/q = (2n)/(n+2)
1 - 1/(p+1) + 1 + 1/q = 2 - 4/(n+2)
1/(p+1) - 1/q = 4/(n+2)
(q-p-1)/[q*(p+1)] = 4/(n+2)
Given that all of p, q, and n are positive integers, then for the last equation to make sense we must have q>p+1.
Now I'll focus just on the left side (q-p-1)/[q*(p+1)]. q is prime so it must be coprime to p+1. Then the difference q-(p+1) is then coprime to p+1 and coprime to q. Since everything is pairwise coprime then q-p-1 is coprime to q*(p+1). But then this means that the fraction (q-p-1)/[q*(p+1)] is irreducible.
Then the reduced form of 4/(n+2) is (q-p-1)/[q*(p+1)], which then means that q-p-1 is a factor of 4. The factors of 4 are 1, 2 and 4. Then there are three possibilities: q-p-1=1, q-p-1=2, or q-p-1=4.
From these we get q-p=2, q-p=3, or q-p=5 as possible differences.
So then only need to show that each possible value of q-p is realized for some pair of primes.
If q-p=2 then q=5 and p=3 yields 3/(3+1) + (5+1)/5 = 39/20 = 2*78/(78+2), which makes n=78
If q-p=3 then q=5 and p=2 yields 2/(2+1) + (5+1)/5 = 28/15 = 2*28/(28+2), which makes n=28
If q-p=5 then q=7 and p=2 yields 2/(2+1) + (7+1)/7 = 38/21 = 2*19/(19+2), which makes n=19
Since everything checks out we can say that all possible values of q-p which satisfy the problem constraints are 2, 3, and 5.
Solution
