1/a + 1/b = 5/391

Change the left-hand side to a single fraction

(a+b)/ab = 5/391

Cross-multiply

391a+391b = 5ab

Move everything to one side

5ab-391a-391b=0

Multiply everything by 5 (so that the coefficient of ab is a square)

25ab-1955a-1955b=0

Add 391^2 to both sides so that the left-hand side can be factorised

25ab-1955a-1955b+391^2=391^2

Factorise the left-hand side

(5a-391)(5b-391)=391^2

Since a and b are positive integers, each of the factors on the left is one less than a multiple of 5, so we must also split the right-hand side into factors that are one less than a multiple of 5. As there are no even factors, this effectively means factors ending in 9. 391 is 17x23, so 391^2 is 17x17x23x23. Numbers ending in 3, when repeatedly multiplied by themselves, end in 1,3,9,7,1,3,9,7, etc. Likewise numbers ending in 7, when repeatedly multiplied by themselves, end in 1,7,9,3,1,7,9,3, etc, moving the opposite way around the same cycle. So the only factors ending in 9 from 17x17x23x23 are 17x17 and 23x23. These equate to 289 and 529.

Let 5a-391=289 and 5b-391=529.

This gives a=136 and b=184. We could just as acceptably assign the factors the other way round to get a=184 and b=136, but these are the only solutions.

Finally, to check:

1/136+1/184 = 23/3128+17/3128=40/3128=5/391