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

Mathematics Asked on January 5, 2022

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.

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.

Answered by user832828 on January 5, 2022

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

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:

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

Answered by Sam on January 5, 2022

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!

Answered by OlympusHero on January 5, 2022