Here's a trick I learned from one of our excellent answerers (heureka)
Let 3 = z
Let j = z + a
Let k = z + b .....so we have
1 1 1
_____ + ______ = ______ get a common denominator on the left
z + a z + b z
[ z + b ] + [ z + a ] 1
________________ = ____
z^2 + az+ bz + ab z
2z + b + a 1
_________________ = _______ cross-multiply
z^2 + az + bz + ab z
2z^2 + az + bz = z^2 + az + bz + ab
2z^2 = z^2 + ab
z^2 = ab
Since z = 3
Then
ab = 3^2
ab = 9
So....we are looking for all the pairs of positive integers that multiply to 9
So we have
a b
1 9
3 3
9 1
So since j = z + a and k = z + b
Then we have
j k
4 12
6 6
12 4
So.....the sum of all possible values for k = 22
EDIT TO CORRECT AN ERROR....
