For each $m$ as the value of $\lfloor\sqrt{n}\rfloor$, $k=0,1,2$ represent the numbers $m^2$, $m^2+m$, and $m^2+2m$.  The next value, $m^2+3m$, is already greater than $(m+1)^2$, and if $n=m^2+3m$, $\lfloor\sqrt{n}\rfloor$ is greater than $m$, so it is no longer a solution for $\lfloor\sqrt{n}\rfloor = m$.