In the equation $\frac{1}{j} + \frac{1}{k} = \frac{1}{3}$, both $j$ and $k$ are positive integers. What is the sum of all possible values for $k$?
+128475 j, k must both be larger than 3
Let 3 = z
Let j = z + a
Let k = z + b
We have
1/ (z + a) + 1/ (z + b) = 1/z
(z+ a + z +b) / [( z+ a) (z + b)= = 1/z cross-multiply
(2z + a + b) z = (z + a) (z + b) simplify
2z^2 + az + bz = z^2 + az + bz + ab
z^2 = ab
3^2 = 9
Factors of 9 = 1, 3, 9
a b z + a = j z + b = k
1 9 3 + 1 = 4 3 + 9 = 12
3 3 3 + 3 = 6 3 + 3 = 6
9 1 3 + 9 = 12 3 + 1 = 4
