if such$\sqrt{37}+\sqrt{47}<\dfrac{n}{m}<\sqrt{41}+\sqrt{43}$ Find this $m$ minimum let $m,n\in N^{+}$, if such
$$\sqrt{37}+\sqrt{47}<\dfrac{n}{m}<\sqrt{41}+\sqrt{43}$$
Find the $m$ minimum the value 
My try: since 
$$(\sqrt{37}+\sqrt{47})m<n<(\sqrt{43}+\sqrt{41})m$$
then
$$\dfrac{10m}{\sqrt{47}-\sqrt{37}}<n<\dfrac{2m}{\sqrt{43}-\sqrt{41}}$$
(maybe this problem use pell equation?)
then I can't,Thank you very much
 A: Since
$$\sqrt{37}+\sqrt{47}=12.9384....$$
$$\sqrt{41}+\sqrt{43}=12.9605....$$
Write
$$\frac{n}{m}=13-\frac{k}{m}$$ Then
$$13-0.0616< 13-\frac{k}{m}< 13-0.039$$
Hence
$$.0616 > \frac{k}{m} \geq \frac{1}{m}$$
This proves that
$$m \geq \frac{1}{0.0616}=16.23$$
Thus, $m \geq 17$.
For $17$ it is easy to show that $13-\frac{1}{17}$ has the desired property.
P.S. If $m > \frac{1}{b-a}$ then it is trivial to prove that there exists an $n$ so that $a\leq \frac{n}{m} <b$. This simple result, shows that any $m \geq 45$ works, and reduces the problem to a finite computation: check which $1 \leq m \leq 44$ works...
Edit Fixed some mistakes in the computations...
A: This provides only a partial answer. What you have gives us
$$\dfrac{\sqrt{43}-\sqrt{41}}2n < m < \dfrac{\sqrt{47}-\sqrt{37}}{10}n$$
Hence, a sufficient condition is that if we ensure that $\dfrac{\sqrt{47}-\sqrt{37}}{10}n -\dfrac{\sqrt{43}-\sqrt{41}}2n > 1$, there is definitely an integer $m$. Hence, $n \geq 7573$. Choosing $n=7573$, gives $m=585$. However, this does not ensure that $m$ is a minimum. All we can say is that the $m$ we are after is $\leq 585$.
