$p$ is a prime iff there exist a unique $m, n\in \mathbb{N}$ such that $\frac{1}{p}=\frac{1}{m}-\frac{1}{n}$

Question

I got this question in one of the whatsapp group I am in:

$p$ is a prime iff there exist a unique $m, n\in \mathbb{N}$ such that $$\frac{1}{p}=\frac{1}{m}-\frac{1}{n}$$

Does this question even make sense.

I can pick $m=2, n=3$ then I get $p=6$ but it is not prime. Is this result true?

Answer 1

Suppose that $p$ is prime. Then by assumtion we have $$(n+p)(p-m)=p^2$$ and we obtain the following cases: $$n+p=p^2$$ and $$p-m=1,$$ which gives $$(m,n)=(p-1,p^2-p)$$ or $$n+p=p$$ and $$p-m=p,$$ which is impossible or $$n+p=1$$ and $$p-m=p^2,$$ which is impossible again.

Id est, we got an unique solution.

Also, by the same way, we see that if $p$ is not prime so we get more systems and we don't get an unique naturals $m$ and $n$.

Answer 2

$$\dfrac12 - \dfrac1{3}=\dfrac16$$ $$\dfrac13 - \dfrac1{6}=\dfrac16$$ $$\dfrac14 - \dfrac1{12}=\dfrac16$$ $$\dfrac15 - \dfrac1{30}=\dfrac16$$

so there is no unique representation and $6$ is not prime

Added

When $p$ is prime the unique representation is $$\dfrac1{p-1} - \dfrac1{p(p-1)}=\dfrac1p$$

If $p=ab$ with both $a>1$ and $b>1$, then you will at least also have $$\dfrac1{a(b-1)} - \dfrac1{ab(b-1)}=\dfrac1{ab}=\dfrac1p$$