Question, help quick

Oops, I meant $k$, not $K$. And positive, not positize. Sorry.

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....

Sorry, the answer was incorrect, here's the solution:

$\text{Multiplying both sides of the equation by $3jk$ to clear the denominator gives $3k + 3j = jk$. Re-arranging and applying Simon's Favorite Factoring Trick, it follows tha t $$jk - 3j - 3k + 9 = (j-3)(k-3) = 9.$$ Thus, $j-3$ and $k-3$ are pairs of positive factors of $9$, so $(j-3,k-3) = (1,9),(3,3),(9,1)$. These give $k = 4,6,12$, and their sum is $4 + 6 + 12 = \boxed{22}$.}$