How to turn arbitrary fractions into arbitrary egyptian fractions?

Question

I am reading Stillwell's Numbers and Geometry. There is an exercise about Egyptian fractions which is the following:

I've tried to do it in the following way - Expressing an arbitrary fraction $\frac{n}{m}$ as the sum of two aribtrary egipcian fractions:

$$\frac{1}{a}+\frac{1}{b}=\frac{n}{m}$$

$$\frac{1}{a}+\frac{1}{b}-\frac{n}{m}=0$$

$$\frac{bm+am-abn}{abm}=0$$

Then solving for $a$ yields:

$$a=\frac{b m}{b n-m}$$

And then expressing it as a function:

$$f(b)= \frac{b m}{b n-m}$$

I have tested with $\frac{n}{m}=\frac{3}{4}$ on Mathematica and all the sums of unit fractions I've obtained equals $\frac{3}{4}$. Is this correct? I'm afraid there's something wrong and I'm not seeing it.

Answer 1

Here's one systematic method to see if there are ways of expressing a fraction as a sum of two Egyptian fractions:

$$\begin{align}\frac{4}{5}=\frac{1}{m}+\frac{1}{n} &\Leftrightarrow 4mn=5m+5n \\ &\Leftrightarrow 5m-4mn+5n=0 \\ &\Leftrightarrow \left(2m-\frac{5}{2}\right)\left(\frac{5}{2}-2n\right)=-\frac{25}{4}\\ &\Leftrightarrow (4m-5)(5-4n)=-25\end{align}$$

Notice that $m,n$ are integers, and so $(4m-5),(5-4n)$ are integers too. Therefore, those have to be divisors of $25$ and there are finitely many divisors of $25$, allowing you to test all possible combinations.

You can use a similar method for three fractions (if no representation using two fractions is possible), but as it has been pointed out in the comments already, it's better to test some small fractions until you get what you want.

Answer 2

As I said a decision for 3 terms there. Erdős-Straus conjecture

As for the answers of two terms, the decision is not always exist.

For the equation: $$\frac{1}{X}+\frac{1}{Y}=\frac{b}{A}$$

You can write a simple solution if the number on the decomposition factors as follows: $$A=(k-t)(k+t)$$
then: $$X=\frac{2k(k+t)}{b}$$ $$Y=\frac{2k(k-t)}{b}$$

or: $$X=\frac{2t(k-t)}{b}$$ $$Y=\frac{-2t(k+t)}{b}$$

First decide so $b=1$ and then I look at what other values have solutions.