jasperE3 wrote: Assume $f:\mathbb N\to\mathbb N$ is a function such that $f(1)>0$ and, for any natural numbers $m$ and $n$, $$f\left(m^2+n^2\right)=f(m)^2+f(n)^2.$$(a) Calculate $f(k)$ for $0\le k\le12$. (b) Calculate $f(n)$ for any natural number $n$. Not gonna lie im kinda confused because of the $f(1)>0$... How is $\mathbb{N}$ defined?

Here, $0\in\mathbb N$, so it is similar to other already-posted problems.

jasperE3 wrote: Here, $0\in\mathbb N$, so it is similar to other already-posted problems. indeed It is definitely possible to bash over $\mathbb{N}$ too actually, using Brahmagupta's identity sufficiently many times, anyone wanna compete for least substitutions?

Consider the similar form with the Cauchy‘s equation,I think it's possible to use the conclusion. When m=0,n=0, we have f(0)=2f(0)^2, so f(0)=0 When n=0, f(m^2)=f(m)^2 so if we let x=m^2, y=n^2 (x,y belongs to Z≥0) f(x+y)=f(x)+f(y). According to the Cauchy's equation, f(x)=kx. f(1)>0, so k>0.f(x)=kx (k is a positive number)

But it's Cauchy's equation over $\{x^2|x\in\mathbb N_0\}$, which has nonlinear solutions.

We take m = n = 0 , which gives us f(0)=0 . Using this fact what we get is f(m^2) = f(m)^2 . From this , it is evident that f(1) = f(1)^2 = 1 . Doing similar computations we get the list of results f(2) = 2 , f(3)=3 , f(4)=4 , f(5) = 5 , f(8)=8 , f(10)=10 , f(9)=9 , f(6) = 6 , f(7)=7 . [These all follows from from some simple calculations] So we derived that for all n <= 10 we get f(n)=n . Now we apply these useful identities below and induction to conclude the problem and see that f(n)=n: (5k+1)^2 + 2^2 = (4k+2)^2 + (3k-1)^2 (5k+2)^2 + 1 = (4k + 1)^2 + (3k+2)^2 (5k+3)^2 + 1 = (4k+3)^2 + (3k+1)^2 (5k+4)^2 + 2^2 = (4k+2)^2 + (3k + 4)^2 (5k+5)^2 = (4k + 4)^2 + (3k+3)^2 . Hope this helps

Got another idea. NOW I would like to prove $f(n)=n$. We define the set $S:=\{n \in \mathbb N_0|f(n)=n\} $. Obviously$ 0,1\in S$. IF 2of $m,\ n,\ m^2+n^2$ belongs to $S$,then the third of it must belongs to $S$. As a result. IF $a^2+b^2=c^2+d^2$, 3of a,b,c,d belongs to S,then the fourth of it must belongs to $S$ IF $S\neq \mathbb N_0$ Assume all $m<n$ belongs to $S$，n does not belong to S IF n is odd. $n=2m+1$ we know that $1,3,5\in S$. $n^2+(m-2)^2=(m+2)^2+(2m-1)^2. m-2,m+2,3m-1<n$, n belongs to S IF n is even. $n=2m$ we know that $2,4,6\in S$.$n^2+(m-5)^2=(2m-4)^2+(m+3)^2$. similarly, we get n belongs to S. SO $S=\mathbb N_0$

What is this $\mathbb{N}_0$? Which numbers does it include?

$\mathbb N_0=\mathbb N\cup\{0\}=\{0,1,\ldots\}$

Wait, a similar problem was in JMO 2021. Right? Just the domain didn't include $0$.

That would have been $f(m^2+n^2)=f(m)f(n)$.