Prove a sufficient condition for $\frac{n}{p}$ to have an egyptian fraction expansion of length $2$.

Question

Prove that $\frac{n}{p}$ has an egyptian fraction expansion of length $2$ if and only if $n|(p+1)$ where $p$ is a odd prime and $n<p$.

Answer 1

Extended HINT: The following examples should suggest an approach to proving that if $n\mid p+1$, then $\frac{n}p$ can be written as the sum of two Egyptian fractions:

$$\begin{align*} \frac2{11}&=\frac16+\frac1{66}\qquad\qquad\frac3{11}=\frac14+\frac1{44}\\\\ \frac4{11}&=\frac13+\frac1{33}\qquad\qquad\frac6{11}=\frac12+\frac1{22}\;. \end{align*}$$

For the other direction, suppose that

$$\frac{n}p=\frac1a+\frac1b=\frac{a+b}{ab}\;;$$

clearly $p\mid ab$, so without loss of generality we may assume that $a=pr$ for some $r\in\Bbb Z^+$, so that

$$\frac{n}p=\frac1{pr}+\frac1b=\frac{pr+b}{pbr}\;.$$

It follows that $brn=pr+b$ and hence that $b=rs$ for some $s\in\Bbb Z^+$. See if you can complete the proof from here. If you get completely stuck, I’ve done most of the rest in the spoiler-protected block below; mouse-over to see it.

Then $\displaystyle\frac{n}p=\frac{pr+rs}{pr^2s}=\frac{p+s}{prs}$, and hence $\displaystyle n=\frac{p+s}{rs}$. But then $s\mid p$, so $s=p$ or $s=1$.