Prove that $a|p+1$ if and only if ${\frac{a}{p} = \frac{1}{n} + \frac{1}{m} }$ such that $n,m \in N$ and $p$ is prime

Question

Let $p$ be a prime and let $n$ be a natural number such that $n \gt a$ .

Prove that $a|p+1$ if and only if exists integers $n,m$

Such that ${\frac{a}{p} = \frac{1}{n} + \frac{1}{m}}$

Answer 1

One direction:

Say $${a\over p}= {1\over m}+{1\over n}\implies amn = p(m+n)$$

Let $g= \gcd(m,n)$ then exists $x,y$ relatively prime such that $m= gx$ and $n= gy$. Then we have $$axyg = p(x+y)\implies x\mid p(x+y) \implies x\mid py \implies x\mid p $$

Case 1: $x=p$ then $ayg = p+y \implies y\mid p\implies y= 1 \implies a\mid p+1$

Case 2: $x=1$ then $ayg = p(1+y) \implies y=1\; \text {or}\; y=p$

If $y=1$ then $m=n$ and $2p= an$ then we can take $n=p$ and $a=2$ and the statement is not true!

If $y=p$ then we get $a\mid p+1$ and we are done.

Other direction: $a\mid p+1$ then $p+1 =an$ so ${a\over p} = {1\over p} +{1\over pn}$ and we are done.

Answer 2

$a\mid p+1 \iff a\cdot n=p+1\\ \iff\frac{a\cdot n}{p}=1+\frac{1}{p} \\ \iff\frac{a}{p}=\frac{1}{n}+\frac{1}{n\cdot p}$

Let $m:=n\cdot p$ and we are done