Functional Equation along the Rationals

Let $x=\tfrac mn$ and define $d=\gcd(m-1,n-1)$ . Then $f\left(\dfrac{x-f(x)}{1-f(x)}\right)=f\left(\dfrac{\tfrac mn-\tfrac 1n}{1-\tfrac 1n}\right)=f\left(\dfrac{m-1}{n-1}\right)=\dfrac d{n-1}.$ Hence the equation in question becomes $\frac d{n-1}=\frac 1n+\frac9{52}$ . Multiplying through to clear denominators gives $\begin{aligned}52nd&=52(n-1)+9n(n-1)\\&=9n^2+43n-52\\\implies 52d&=9n+43-\dfrac{52}n.\end{aligned}$ Thus $n|52$ , so the possible values of $n$ are $1,2,4,13,26,52$ . Testing all of these gives $(n,d)=(13,3)$ as the only solution. Hence the set of possible $m$ is the set of integers with $1\leq m\leq 12$ that satisfy $\gcd(m-1,12)=3$ , of which there are only two: $4$ and $10$ . Therefore the only two rational numbers $0<x<1$ that satisfy this equation are $x=\tfrac 4{13}$ and $x=\tfrac{10}{13}$ . The product of these two numbers is $\tfrac{40}{169}$ and the desired sum is $40+169=\boxed{209}$ .

Let LHS = 1/a and then let f(x) = 1/b. We now get 1/a = 1/b + 9/52. By Rearranging we get a/52 = b/(52-9b) we need to only test b = 1, 2,3,4 and 5 as RHS in this new expression is positive. From this we get that b =4 and a = 13 is the only (a,b). Now we let x = c/13 for 1=<c<=12 so that the new RHS is 1/4. If we let x = c/13 then in the LHS of the original expression we have f((c-1)/12). Now (c-1) must have 3 as one of its factors and not 2 since the denominator is (3)(4) and LHS must be 1/4. In the interval this is the case when c-1=3 and when c-1=9. Thus there are 2 solutions for c which are 4 and 10. Now (4/13)(10/13) = 40/169. 40 + 169 = 209