# $2f(m^2 + n^2) = f(m)^2 + f(n)^2$

Determine all $$f : \mathbb N_0 \to \mathbb N_0$$ that satisfies $$2f(m^2 + n^2) = f(m)^2 + f(n)^2$$ and $$f(m^2) \geqslant f(n^2)$$ 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!

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.

• A small typo: $f(0^2+0^2) = 2f(0)^2$. – PierreCarre May 22 '20 at 15:08
• Wouldn't the second point suggest that $f(0)$ is either 0 or 2? – For the love of maths May 22 '20 at 15:35
• True. I will edit or somebody please edit the post – dust05 May 24 '20 at 2:43