$f:\mathbb N_0\to\mathbb N_0$ with $2f\left(m^2+n^2\right)=f(m)^2+f(n)^2$ and $f\left(m^2\right)\geqslant f\left(n^2\right)$ when $m\geqslant n$ 
Determine all $f : \mathbb N_0 \to \mathbb N_0$ that satisfy $2f\left(m^2 + n^2\right) = f(m)^2 + f(n)^2$ and $f\left(m^2\right) \geqslant f\left(n^2\right)$ when $m \geqslant n$.

I managed to prove that $f$ is increasing, but I don't know what to do next. Can anyone give some hints or solution please. Thank you very much!
 A: Observe the following things : 


*

*For any $m, n \in \mathbb{N}_0$, $f(m)^2 + f(n)^2 $ is an even number, so the values of $f(x)$ are all odd or all even. 

*$2f(0^2+ 0^2) = 2f(0)^2$ so we have $f(0) = 0$ or $1$.

*$2f(1^2 + 0^2) = f(1)^2 + f(0)$. 
First assume that $f(x)$ values are all odd. then we have $f(0 ) = 1$ and  $(f(1)-1)^2 = 0$, $f(1) = 1$. Again $f(2) = f(1)^2 = 1$. If $f(a) = 1$, then $2f(a^2) =f(a)^2 + f(0) = 2$, i.e. $f(a^2) = 1$. This leads us to $f(2) = f(4) = f(16) = \cdots = 1$, and from the fact that $f$ is increasing (I didn't check this but you said you proved this, so, ) we can see that $f(x) = 1$ for all $x \in \mathbb{N}_0$. 
For the second case we assume that $f(x)$ values are all even. $f(0) = 0$, and $f(1) = 0$ or $2$. If $f(1) = 0$, we have $f(2) = 0$ and $f(4) = f(16) = \cdots = 0$. One can verify this as same as $f$ of odd values case. 
The only remaining case is that $f$ has even values, $f(0) = 0$, $f(1) =2$.
From now on, for simplicity, Let $g(x) = f(x)/2$. Then $g$ is also an integer function and satisfies $g(n^2+m^2) = g(n)^2 + g(m)^2$. From this step, we can refer to this paper. I think the additional conditions we have would make this problem much easier, but I will proceed after the paper. 
Note that $(xy + zw)^2 + (xw-yz)^2 = (xy - zw)^2 + (xw + yz)^2$, so $g(xy + zw)^2 + g(xw-yz)^2 = g(xy - zw)^2 + g(xw + yz)^2$ whenever the terms are nonnegative. 
Putting $(x, y, z, w) = (k, 2, 1, 1)$ lead us to 
$$ g(2k+1)^2 = g(2k-1)^2 + g(k+2)^2 -  g(k-2)^2.$$ If $g(x) = x$ for $x \le 2k-1$, one can wee that $g(2k+1) = (2k-1)^2 +(k+2)^2 - (k-2)^2 = 4k^2  + 1+ 4k  = (2k+1)^2$, i.e. $g(2k + 1) = 2k+ 1$. 
Putting $(x, y, z, 2) = (k-1, 2, 2, 1)$ lead us to the similar induction step for even $x$ case. 
So, to show that $g(x) = x$, i.e. $f(x) = 2x$, it is enough to show for some initial cases. Refer to the paper I linked for remaining details. 
