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 $0 < n < m$.
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 $0 < n < m$.
I assume r to be irrational, else there will be no non-trivial solution.
So, $(m-n)n=\frac{2r}{\sqrt3}$
Now $m-n
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.
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.
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
Not sure it will really help...