# Find two fractions such that their sum added to their product equals $1$

This is a very interesting word problem that I came across in an old textbook of mine. So I managed to make a formula redefining the question, but other than that, the textbook gave no hints really and I'm really not sure about how to approach it. Any guidance hints or help would be truly greatly appreciated. Thanks in advance :) So anyway, here the problem goes:

Find two fractions so that their sum added to their product equals $1$.

In other words:

Given integers $a$ and $b$ with $\frac ab < 1,$ find integers $c$ and $d$ such that $\frac ab + \frac cd + \frac ab \cdot\frac cd = 1$

• Finding $\frac{c}{d}$ to given $\frac{a}{b}$ may be much harder, that finding any pair $\frac{a}{b}, \frac{c}{d}$ such that $\frac ab + \frac cd + \frac ab \cdot\frac cd = 1$. – Tacet May 27 '16 at 10:27
• The paradigm is $\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1$ – almagest May 27 '16 at 10:29

$$\frac ab + \frac cd + \frac ab \cdot\frac cd = 1\implies\frac ab+\left(1+\frac ab\right)\frac cd=1$$ Solve for $\frac cd$.

$$x+y+xy = 1 \quad \Longleftrightarrow\quad (x+1)(y+1)=2$$ hence assuming $x=\frac{p}{q}$, $y=\frac{q-p}{q+p}$ does the job.

Simple solution may be $(a,b,c,d) = (1,1,0,1)$.

$$\frac ab + \frac cd + \frac ab \cdot\frac cd = 1 \Leftrightarrow \frac{ad+cb+ac}{bd}=1 \Leftrightarrow\\ a(d+c) = b(d-c)\Leftrightarrow \frac{a}{b} = \frac{d - c}{d+c} \$$

Take some $d,c$. It has to satisfy $d \neq 0 \wedge d+c \neq 0$. Eg. $(d,c) = (7,3)$ then $\frac{a}{b} = \frac{d-c}{d+c} = \frac{2}{5}$. Possible (example) solution is $(a,b,c,d) = (2,5,3,7)$, you can check it.
Given integers a and b with $ab<1$, find integers c and d such that $\frac ab + \frac cd + \frac ab \cdot\frac cd = 1$
You have to solve $$\begin{cases} a = d - c\\ b = d + c\\ \end{cases}$$