$a\in \mathbb{N}$, $p$ prime, $a<p$ prove that $a\mid p+1\iff\exists\, b,c\in\Bbb N:\dfrac{a}{p}=\dfrac{1}{b}+\dfrac{1}{c}$ [duplicate]

$a\in \mathbb{N}$, $p$ prime, $a<p$ prove that $a\mid p+1\iff \exists\, b,c\in\Bbb N:\dfrac{a}{p}=\dfrac{1}{b}+\dfrac{1}{c}$

my attempt:

$a\cdot b \cdot c=p\cdot(b+c)$ .

i dont know how to use the given
$a\mid p+1$

Solution 1:

We have $ a \mid p+1$ so there is $\lambda$ such that $\lambda a =p+1$. Now divide by $ \lambda p$ & we have \begin{eqnarray*} \frac{a}{p} = \frac{1}{\lambda} + \frac{1}{p \lambda}. \end{eqnarray*}

The other implication: We have $ \dfrac{a}{p}=\dfrac{1}{b}+\dfrac{1}{c}$ or $abc=p(b+c)$.Multiply this by $a$ & rearrange to $(ab-p)(ac-p)=p^2$.

This gives three possiblities $ab-p=1$ or $ac-p=1$ & the result follows. Or $ab-p=p,ac-p=p$ which gives $ab=ac=2p$ so $a=1$ or $a=2$ and again the result follows.