Finding integer solutions to $m$ and $n$ How can we find the positive integer solutions to the variables $m$ and $n$, if we know $r$, that satisfy the equation:
$$r = \frac{\sqrt{3(m-n)^2 n^2}}{2},$$
where $m$ and $n$ are coprime, and $0 < n < m$.
 A: Hint: If $r$ is rational, hardly any. Square both sides, simplify a bit. The $3$ kills us except if $m=n$. 
Edit: With the newly added restriction $0\lt n \lt m$ there are no solutions. 
A: Let's suppose $r\ge 0$ (because of the square root) then squaring we get :
$$\tag{1} 4r^2=3(m-n)^2n^2$$
If  $r$ is supposed integer then we need $3|r$ i.e. $r=3k$ with $k$ a nonnegative integer (because $m$ and $n$ are integer) and your equation becomes :
$$4\cdot 3k^2=(m-n)^2n^2$$
but this can't have a positive solution since the number of $3$ at the left is odd while the number of $3$ at the right is even.
This implies that $k=0,\ n=0,\ m=0$.

If $r$ is not supposed integer then your equation becomes simply :
$$r'=\frac {2r}{\sqrt{3}}=(m-n)n\quad \text{(since $\ 0<n<m$)}$$
Since we want $m$ and $n$ integer $r'$ must be integer and may be :


*

*$r'=1.n$ (if $m-n=1$ corresponding to the trivial solution $n=r',\ m=r'+1$)

*$r'=p.n$ with $p$ and $n$ coprime ($p=m-n$) : i.e. computing $r'$ you got an integer that can't be power of prime ; decompose it in powers of primes :
$$r'=\prod_{i=1}^N p_i^{k_i}$$
and consider all the partitions in two classes possible of these $N$ primes, one will define $n$ and the other $p$ (after that deduce $m=n+p$).


Not sure it will really help...
A: I assume r to be irrational, else there will be no non-trivial solution.
So, $(m-n)n=\frac{2r}{\sqrt3}$
Now $m-n<m$ and $n<m =>(m-n)n<m^2=>m^2>\frac{2r}{\sqrt3}$
Again  $m>n=>m-n≥1=>n≤\frac{2r}{\sqrt3}$
So, $1≤n≤[\frac{2r}{\sqrt3}]$  where [] is the greatest integer funtcion.
The value of m can be calculated from the given equation. We need to check whether (m,n)=1 is satisfied.
