Tom K. answered 10/21/20
Knowledgeable and Friendly Math and Statistics Tutor
1/m + 1/n = (n+m)/mn
(n+m)/mn = 2/p
Thus, p|mn
As n > m, 1/m > 1/p, so m < p, and p|n
m | n, or the denominator of 1/m + 1/n must contain a factor other than p, which would be a contradiction
Thus, 1/m + 1/cpm = 2/p for some c.
(cp+1)/cpm = 2/p
(cp+1)/cm = 2
2cm = cp + 1
Thus, as c | 2cm and c|cp, c|1, so c = 1
Thus, 1/m + 1/pm = 2/p
(p+1)/pm = 2/p
2m = p+1
m = (p+1)/2
n = p(p+1)/2