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 $0 < n < m$.

  • $\begingroup$ What is the nature of r, integral, rational or ...? $\endgroup$ – lab bhattacharjee Aug 15 '12 at 17:34
  • $\begingroup$ That way $r$ is always irrational number you have a $\sqrt{3}$ there. $\endgroup$ – clark Aug 15 '12 at 17:38
  • $\begingroup$ I don't think $r$ can be an integer anyway, rearranging gives $4r^2 = 3(m-n)^2 n^2$ and there is a problem with divisibility of $3$ on both sides if $r$ is an integer. Or taking clark's even simpler observation... $\endgroup$ – fretty Aug 15 '12 at 17:39
  • $\begingroup$ No, $r$ isn't an integer, the integers here are m and n. However r must be positive and non-zero of course. $\endgroup$ – Khaled Aug 15 '12 at 17:47

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...

  • $\begingroup$ I never said $r$ was an integer. The integers here are $m$ and $n$. $\endgroup$ – Khaled Aug 15 '12 at 18:05
  • $\begingroup$ @Khaled: I'll edit my answer to support real $r$, $\endgroup$ – Raymond Manzoni Aug 15 '12 at 18:07

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.

  • $\begingroup$ Except in trivial cases, which your new edit has ruled out, if the equation you wrote above is true, then $\sqrt{3}$ is rational. But it isn't. The argument has essentially been given in one of the answers. Let $r=a/b$ where $a$ and $b$ are integers. Then $4r^2=4(a/b)^2=3(m-n)^2n^2$, so $4a^2=(3)(b^2)(m-n)^2n^2$. The number of $3$'s in the prime factorization of $4a^2$ is even, but the number of $3$'s in the prime factorization of $(3)(b^2)(m-n)^2n^2$ is odd, so the two numbers can't be equal. $\endgroup$ – André Nicolas Aug 15 '12 at 18:05

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.


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