$k=m\to n=[\sqrt 2 m]$. Let $n=\sqrt 2 m-\theta, 0<\theta<1, l=n^2-2m^2>0, \theta=\frac{l}{n+m\sqrt 2}.$ Then $[\sqrt 2 k-\frac{\theta k}{m}]=[\sqrt 2 k]\to \frac{\theta k}{m}\le \{\sqrt 2 k\}.$ Consider solution of Pell's equation $x^2-2y^2=-1, (x,y)=(1393,985)$. From $1393^2-2*985^2=-1\to 985\sqrt 2=1393+\frac{1}{1393+985\sqrt 2}$. Take $k=985$: $\frac{985l}{m(n+m\sqrt 2)}\le \frac{1}{1393+985\sqrt 2}.$ - non solution for $m<2378$. Minimal solution is $(n.m)=(3263,2378)$ solution of Pell's equation $x^2-2y^2=1,y>2007$.

horizon wrote: find all positive integer pairs $(m,n)$,satisfies: (1)$gcd(m,n)=1$,and $m \le r$ (2)for any $k=1,2,...,r $,we have $[\frac{nk}{m}]=[\sqrt{2}k]$ I think the generalisation where $r$ is some natural number has always at least one value for $m$ as solution in the question. Take the value $m$ in $\{1,2,\cdots, r\}$ such that $\frac{ \lfloor{\sqrt{2}k\rfloor} }{k}$ maximal is. It is easy to see that this value satisfy for each $k \in \{1,2,\cdots, r\}$ *** Which solution of those is correct/ if Rust is correct, it seems I made a big mistake?