Finding perfect two terms Egyptian fraction If I have two unity fractions, like $\frac{1}{12} + \frac{1}{180}$, for instance. These two fractions can be re-writen as $\frac{1}{15} + \frac{1}{45}$ or even $\frac{1}{18} + \frac{1}{30}$, which satisfies the Egyptian Fraction concept of maximising the value of the smallest fraction. Is there a formula or an algorithm that when followed, one can find out all the possible equal pairs of any unity fraction pair?
 A: This might be worth something.
Let $x=p/q, p<q.$
Then $1<q/p=[q/p]+z, z<1$, where [] is the greatest integer function or floor.
Now $[q/p]<q/p\implies p/q <1/[q/p]$, but we need to add additional fractions  $1/[p/q]$ to get $p/q$. This is not possible since the first term already exceeds the fraction. So we have to add 1 to the denominator as we go.
So $p/q=\frac{1}{1+[q/p]}+u/v$.
$u/v=p/q-\frac{1}{1+[q/p]}<1$ and the next term is $\frac{1}{1+[v/u]}$
$p/q=\frac{1}{1+[q/p]}+\frac{1}{1+[v/u]}+...$
This iterative process generates the Egyptian Fractions for a given number.
A: For a given rational number $\dfrac{c}{d}$, we wish to find all pairs of positive integers $(a,b)$ such that $$\dfrac{c}{d} = \dfrac{1}{a}+\dfrac{1}{b}.$$
We can manipulate the equation as follows: $$cab = da+db$$ $$cab - da-db = 0$$ $$c^2ab-cda-cdb = 0$$ $$c^2ab-cda-cdb+d^2 = d^2$$ $$(ca-d)(cb-d) = d^2.$$
Hence, $ca-d$ and $cb-d$ must be complementary factors of $d^2$. So the pairs of positive integers $(a,b)$ such that $\dfrac{c}{d} = \dfrac{1}{a}+\dfrac{1}{b}$ are all of the form $$(a,b) = \left(\dfrac{f_1+d}{c},\dfrac{f_2+d}{c}\right)$$ for all integers $f_1,f_2$ such that $d^2 = f_1f_2$ and both $\dfrac{f_1+d}{c}$ and $\dfrac{f_2+d}{c}$ are integers.
