find all positive integer $m,n$ Given $p$ is prime number, find all pairs of positive integers $m,n$ such that $\gcd(m,n)=1$ and 
$$\frac{p+m}{p+n}-\frac{m}{n}=\frac{1}{p^{2}}$$
We can write this equation as:
$$\frac{p(n-m)}{n(p+n)} = \frac{1}{p^2}$$ 
Since we have positive RHS and $n(p+n)>0$, we also have:
$$p(n-m)>0 \implies n>m$$
I found that $n>m$, but I have no idea to complete. Please give me some ideas, thank you very much!!
 A: We have:
$$\frac{p(n-m)}{n(p+n)}=\frac{1}{p^2} \implies p^3(n-m)=n(p+n)$$
This gives $n \mid p^3(n-m)$. But we know that:
$$\gcd(m,n)=\gcd(n,n-m)=1$$
by Euclidean Algorithm. Thus, we have:
$$n \mid p^3(n-m) \implies n \mid p^3$$
Since there is a positive integer lesser than $n$ (namely $m$), we cannot have $n=1$.
If $n=p$, then:
$$p^3(p-m)=2p^2 \implies p(p-m)=2$$
which gives $p=2$ and $m=1$. Thus, $(m,n,p)=(1,2,2)$ is a solution.
Else if $n=p^2$, then:
$$p^3(p^2-m)=p^2(p^2+p) \implies p^2-m=p+1 \implies m=p^2-p-1$$
which gives $(m,n,p)=(p^2-p-1,p^2,p)$ for any prime $p$.
Else $n=p^3$, then:
$$p^3(p^3-m)=p^3(p^3+p) \implies p^3-m=p^3+p \implies m=-p$$
which is clearly impossible.
Thus, the solutions are:
$$(m,n,p)=(1,2,2),(p^2-p-1,p^2,p)$$
A: Another approach:
We may write:
$$\frac{1}{p^2}=\frac{p(n-m)}{n(p+n)}$$
Clearly $p|n$,  let $n=kp$; following cases can be considered:
1):$k=1$⇒ $n=p$⇒
$$\frac{p(p-m)}{p(p+p}=\frac{p^2-pm}{2p^2}=\frac{1}{p^2}$$
$p(n-m)=p^2-mp=2$ ⇒ $p^2+mp-2=0$
$\Delta=m^2+8$ ; $m=1$ ⇒$ n=p=1$ ⇒$(m, n, p)=(1, 1,1) $, but 1 is not prime so this result is not acceptable.
2):$k=2$ ⇒ $2p^2-mp-6=0$
$m=1$⇒$p=2$ and $n=4$⇒$(m, n, p)=(1, 4, 2)$
3): $k=3$⇒$3p^2-mp-12=0$
$m=5$⇒$p=3$ and $n=9$
Similarly we can find numerous values for m, n and p. We can see that :
$m=1=2^2-2-1$
$m=5=3^2-3-1$
and conclude that:
$m=p^2-p-1$ 
Therefore general form of solutions can be:
$(m. n. p)=(p^2-p-1, n=kp, p)$
