If m, n are positive integers and $m+n\sqrt{2}=\sqrt{41+24\sqrt{2}}$   then $\left(m+n\right)$  is equal to

Squaring both sides, we get

$m^2+2n^2+2\sqrt{2}mn=41+24\sqrt{2}$

On comparing, we get

$m^2+2n^2=41\cdots \left(1\right)$

$\begin{array}{c}mn=12\\ \Rightarrow n=\frac{12}{m}\mathrm{......}\left(2\right)\end{array}$

On solving $\left(1\right),\left(2\right)$ , we get

$\Rightarrow m^2+\frac{2\left(144\right)}{m^2}=41$

$\Rightarrow m^4-41m^2+288=0$

$\Rightarrow \left(m^2-32\right)\left(m^2-9\right)=0$

$\Rightarrow m^2\ne 32,m^2=9$

$\Rightarrow m=3andn=\frac{12}{3}=4$

$\therefore \left(m+n\right)=3+4=7$