How many numbers of the form $m^2 + \sqrt{2} n^2$ are between $1 \times 10^6$ and $2 \times 10^6$? I have a purely computational question today.  How many numbers of the form $m^2 + \sqrt{2} n^2$ with $m,n \in \mathbb{Z}$ are between $1 \times 10^6$ and $2 \times 10^6$ ?
$$ \# \big\{ (m,n):  1 \times 10^6 < m^2 + \sqrt{2} \, n^2 < 2 \times 10^6 \big\} =\;  ? $$
The asymptotic answer for this sequence of numbers can be found using Weyl's law, for exmaple that:
$$ \# \big\{ (m,n) : 0 < m^2 + \sqrt{2} \, n^2 < X \big\} \sim \frac{\pi}{4 \sqrt{2}} \, X$$
This result is saing that numbers of the form $m^2 + \sqrt{2} \, n^2$ are roughtly linearly distributed on the real line $\mathbb{R}$.  
The Weyl's law estimate gives $\frac{\pi}{4\sqrt{2}} \times 10^6 \approx 555,630 + 0.36726\dots$  Perhaps with a computer it's possible to obtain an exact answer?  E.g. using Python or sage.

COMMENT.- Without calculator the problem is 
 tedious but not difficult.
The required points are enclosed between the ellipses of equations $m^2 + \sqrt{2} n^2=10^6$ and $m^2 + \sqrt{2} n^2=2\cdot10^6$. Define the following sets in the first quadrant:
$$A=\{(0,y)\text { with } y\gt0 \}\\B=\{(x,0)\text { with } x\gt0 \}\\C=\{(x,y)\text { with } x\gt0,y\gt0 \}$$ By symmetry we have for the total number $N$ of points
$$N=2A+2B+4C$$ $A$ has $573$ points and $B$ has $413$.
One has for the calculation of the set $C$ $$\frac{\sqrt{10^6-m^2}}{\sqrt[4]2}\lt y\lt\sqrt[4]2\sqrt{10^6-m^2}\text { where } 1\le m\le1000$$   Assuming that @rogerl's comment above is true, there are $4\cdot660074+1972=\color{red}{2\space642\space268}$ points in total.
