Positive integer solutions to $\frac{1}{a} + \frac{1}{b} = \frac{c}{d}$

Question

I was looking at the equation $$\frac{1}{a}+\frac{1}{b} = \frac{c}{d}\,,$$ where $c$ and $d$ are positive integers such that $\gcd(c,d) = 1$.

I was trying to find positive integer solutions to this equation for $a, b$, given any $c$ and $d$ that satisfy the above conditions. I was also trying to find whether there are additional requirements on $c$ and $d$ so that positive integer solutions for $a$ and $b$ can even exist.

I found that this equation simplifies to $abc - ad - bd = 0$ so that $abc = d(a+b)$.

Also, since the equation is equivalent to $a+b = ab(\frac{c}{d})$, this means $a$ and $b$ are the roots of the quadratic $dx^2-abcx+abd = 0$ since their product is $ab$ and their sum is $a+b = ab(\frac{c}{d})$.

However, after I analyzed the quadratic I just ended up with $a = a$ and $b = b$.

Any ideas on how to solve this further?

Again, I need to find all the conditions on the positive integers $c$ and $d$ (where $\gcd(c,d) = 1$) such that positive integer solutions for $a, b$ can exist. And then also find the positive integer solutions for $a$ and $b$ given that those conditions are satisfied.

Answer 1

Try some examples. I suggest researching Simon's Favorite Factoring trick, it helps with these types of questions. Asking on artofproblemsolving.com is a great idea, Simon is a real person who invented this factoring trick, and it's on that website.

I'll give you one example. Let's take $\frac{1}{a}+\frac{1}{b}=\frac{1}{4}$. Note that $\frac{a+b}{ab}=\frac{1}{4}$. So, $4a+4b=ab$, meaning that $ab-4a-4b=0$, so $(a-4)(b-4)=16$. To solve this in the integers, just find the factor pairs of $16$ and solve for $a,b$

Hope this helped you!

Answer 2

$\frac{1}{a}+\frac{1}{b} = \frac{c}{d}\,,$

Above has solution:

$a=(3k-2)$

$b=(k-2)(3k-2)$

$c=(k-1)(3k-2)$

$d=(k-2)(3k-2)^2$

For k=5 we get:

$(a,b,c,d)=(13,39,52,507)$

Answer 3

The first equation is equivalent to (ac-d)(bc-d) = d^2. From there, you can find all the ways that two numbers multiply to d^2. If you have two numbers e and f that multiply to be d^2, then you can solve ac-d = e, bc-d = f individually. This stems from Simon's Favorite Factoring Trick that another person mentioned.