Real solution to the system $\frac{2}{m} + \frac{3}{n} = \frac{1}{p}, m^2 + n^2 = 1360, m + n + p = 47$ A friend of mine gave me the following problem.

Let $m,n,$ and $p$ be the real solution to the system\begin{align*}
 \frac{2}{m} + \frac{3}{n} &= \frac{1}{p}\\ m^2 + n^2 &= 1360\\ m + n +
 p &= 47. \end{align*} What is the value of $n - m$?

My friend got this problem in his test, which should only take around one minute to solve each problem. We know that the solution is $m = 8, n = 36, p = 3$. Well, we used WolframAlpha to find it, to be honest (not on the test, of course!). But I think there might be no nice solution to this problem, let alone a solution that solves the problem in a short amount of time. And we think it might be that "trap" problem that wastes your time, precious time.
I have tried to solve it without WolframAlpha, though. After getting rid of $p$ and simplifying I got the system
\begin{align}
m^2 + n^2 &= 1360\tag{1}\\
3 m^2 + 6 m n - 141 m + 2 n^2 - 94 n & = 0\tag{2}
\end{align}
See that? Equation $(2)$ is a mess. This is why I expect there is no "nice" solution. (I don't even find any solution yet, sad)
Another way that come to mind was by finding $n - m$ directly, without finding the solution itself. But this also doesn't seem to work.
Any hint? Any solution is welcome.
 A: Assuming that integers will work:
$1360 = 16 \cdot 5 \cdot 17$  If a sum of two squares is divisible by 4, each argument is divisible by 2. This happens twice. We reach  $m = 4 x, n = 4 y$  with $x^2 + y^2 = 85$.
As $85$  is composite, we get two pair, $85 = 81 + 4$ or $85 = 49 + 36.$  The pairs (9,2) and (7,6) come from  that multiplier formula and $5 = 4+1$  and $17 = 16 + 1,$  combined in two orderings...
The two ordered positive $(m,n) = (36,8) \; ; \; (28,24).$  Or $(m,n) = (8,36) \; ; \; (24,28).$  Or placing some minus signs
A: Solutions can be found by testing
$\qquad n=\sqrt{1360-m^2}, \quad
 1\le m \le \lfloor\sqrt{1360}\space \rfloor =36\qquad$
to see which $\space m$-values yield integers. These solutions may then be tested by
$\space p=47-m-n\space$
to see which yield primes and
$\quad p=\dfrac{mn}{2n+3m}\space$ to see which yield integers.  Solutions like
$\space (24,26,\space (28(24)\space (36,8)\space$ yield primes in the subtraction test but none yield integers in the rational test. As it happens, only the first solution $\space (8,36)\space$ meets all criteria.
\begin{align*}
 y&=\sqrt{1360-8^2}=36\\
 p&=47-8-36=3 \\
 p&=\dfrac{8(36)}{2(36)+3(8)}=3
\end{align*}
\begin{align*}
\frac{2}{m}+\frac{3}{n}=\frac{1}{p}\\
\text{The only integer solutions are} \\
\\  
\frac{2}{8}+\frac{3}{36}=\frac{1}{3}
\\
m^2+n^2=8^2+36^2=1360
\\
m+n+p = 8+36+3=47
\\
n=m - 36-8=28
\end{align*}
