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Find all functions $f: \mathbb{N_0} \to \mathbb{N_0} $ such that $$f(f(m)^2+f(n)^2)=m^2+n^2.$$

My work: The given condition implies that $f$ is strictly increasing and injective. Indeed, if $$f(m)=f(n) \implies f(m)^2=f(n)^2 \implies f(m)^2+f(n)^2=f(n)^2+f(n)^2.$$ On applying the function we see that $m=n$. Thus $f$ is injective.

Although I claim that $f$ is strictly increasing, I have not found a convincing proof of that yet. So please help me there, too.

Setting $m=n=0$ in the original equation, we obtain $f(2f(0)^2)=0$. Suppose $f(0)=k > 0$. Then we have $f(2k^2)=0$; i.e. there is an element in the range of $f$ whose $f$ value is smaller than $f(0)$. $$\therefore f(2k^2) < f(0).$$ As $f$ is strictly increasing by the un-proven claim, $$ 2k^2 < 0 \implies k^2 < 0,$$ which is absurd.

$$\therefore k=0 \implies f(0)=0.$$Therefore, $f(f(m)^2)=m^2$. Therefore, $$f(f(m)^2+f(n)^2)=m^2+n^2 \\\implies f(f(m)^2+f(n)^2)=f(f(m)^2)+f(f(n)^2).$$On applying $f$ again we see that $f(m^2+n^2)= f(m)^2+f(n)^2.$ ( I had to look at the hints for this part, sorry.) Then setting $m=1, n=0$ we have either $f(1)=1 $ or $f(1)=0$. If $f(1)=0$ we obtain $0=1$ from the relation and $f(0)=0$. Therefore $f(1)=1$.

A simple inductive step fixes $$f(n)=n. \quad \forall n\in \mathbb{N}$$

I still need the part that fixes $f$ is nondecreasing. Please help and check if the proof is correct.

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    $\begingroup$ If I remember it correctly, this question is proved by induction on $k$ that $$f(8k) = 8k, \cdots, f(8k + 7) = 8k + 7.$$ $\endgroup$ Commented Feb 14, 2018 at 13:00

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An alternative proof of $f(0) = 0$ that doesn't rely on the unproven claim. Say $f(0) = c$. Then:

$$f(f(0)^2 + f(0)^2) = 0$$ $$f(2c^2) = 0$$

But this implies:

$$f(f(2c^2)^2 + f(2c^2)^2) = 8c^4$$ $$f(0) = 8c^4$$

Since we know $c = 8c^4$, we have established that $c$ must be $0$.

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  • $\begingroup$ Thanks! This fixes my answer altogether. $\endgroup$
    – QFTheorist
    Commented Feb 14, 2018 at 12:50

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