Find all $f$ such that $f\left(m^{2}+n^{2}\right)=f(m)^{2}+f(n)^{2},$

Question

Question -

Find all $f: \mathbb{N}_{0} \rightarrow \mathbb{N}_{0}$ which satisfy

(a) $f\left(m^{2}+n^{2}\right)=f(m)^{2}+f(n)^{2},$ for all $m, n$ in $\mathbb{N}_{0}$

(b) $f(1)>0$

my try -

I showed that $f(n)=n$ for all $n<10$ by using given conditions very easily ... but i am not able to apply induction to prove that $f(n)=n$ for all n because there is square term inside...i think this question has different strategy to solve..

any help will be appreciated thankyou

Answer 1

Hint. Induction works. Use the identities $$(2k+1)^2+(k-2)^2=(k+2)^2+(2k-1)^2$$ and $$(2k+2)^2+(k-4)^2=(k+4)^2+(2k-2)^2\,.$$

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By the OP's claim, $f(n)=n$ for $n<10$. Suppose now that $n\geq 10$ is such that $f(m)=m$ for all integers $m$ such that $0\leq m <n$.

If $n$ is odd, then $n=2k+1$ for some integer $k\geq 4$. Therefore, $$\begin{align}\big(f(2k+1)\big)^2+\big(f(k-2)\big)^2&=f\big((2k+1)^2+(k-2)^2\big)\\&=f\big((k+2)^2+(2k-1)^2\big)\\&=\big(f(k+2)\big)^2+\big(f(2k-1)\big)^2\,.\end{align}$$ By induction hypothesis, $f(k-2)=k-2$, $f(k+2)=k+2$, and $f(2k-1)=2k-1$. This gives $f(n)=f(2k+1)=2k+1=n$.

If $n$ is even, then $n=2k+2$ for some integer $k\geq 4$. Apply the same strategy as the previous paragraph to show that $f(n)=n$.

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Anyway, we can easily check that $f(n)=n$ for all $n=0,1,2,\ldots,9$. Plugging $m:=0$ and $n:=0$ into the functional equation shows that $2\,\big(f(0)\big)^2=f(0)$, so $\big(2\,f(0)-1\big)\,f(0)=0$. Since $f(0)$ is an integer, $2\,f(0)-1$ is odd, whence unequal to $0$. This means $f(0)=0$.

Plugging in $m:=1$ and $n:=0$ into the functional equation yields $f(1)\,\big(f(1)-1\big)=0$. As $f(1)>0$, we conclude $f(1)=1$. Hence, $$f(2)=f(1^2+1^2)=\big(f(1)\big)^2+\big(f(1)\big)^2=1^2+1^2=2\,.$$

This shows that $$f(4)=f(0^2+2^2)=\big(f(0)\big)^2+\big(f(2)\big)^2=0^2+2^2=4\,,$$ $$f(5)=f(1^2+2^2)=\big(f(1)\big)^2+\big(f(2)\big)^2=1^2+2^2=5\,,$$ and $$f(8)=f(2^2+2^2)=\big(f(2)\big)^2+\big(f(2)\big)^2=2^2+2^2=8\,.$$ Thus, $$\begin{align} 25&=0^2+5^2=\big(f(0)\big)^2+\big(f(5)\big)^2=f(0^2+5^2)\\ &=f(25)=f(3^2+4^2)=\big(f(3)\big)^2+\big(f(4)\big)^2\\ &=\big(f(3)\big)^2+4^2=\big(f(3)\big)^2+16\,. \end{align}$$ As $f(3)\in\mathbb{Z}_{\geq 0}$, we obtain $f(3)=3$. Consequently, $$f(9)=f(0^2+3^2)=\big(f(0)\big)^2+\big(f(3)\big)^2=0^2+3^2=9\,,$$

Now, $$\begin{align} 50&=5^2+5^2=\big(f(5)\big)^2+\big(f(5)\big)^2=f(5^2+5^2)\\ &=f(50)=f(1^2+7^2)=\big(f(1)\big)^2+\big(f(7)\big)^2\\ &=1^2+\big(f(7)\big)^2=1+\big(f(7)\big)^2\,. \end{align}$$ Thus, $f(7)=7$. Finally, from $$\begin{align} 85&=2^2+9^2=\big(f(2)\big)^2+\big(f(9)\big)^2=f(2^2+9^2)\\ &=f(85)=f(6^2+7^2)=\big(f(6)\big)^2+\big(f(7)\big)^2\\ &=\big(f(6)\big)^2+7^2=\big(f(6)\big)^2+49\,, \end{align}$$ we obtain $f(6)=6$.

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Let $S$ be the subset of $\mathbb{Z}_{\geq 0}$ consisting of all nonnegative integers that can be written as a sum of two perfect squares of integers. For a description of $S$, see this link. The solutions $f:\mathbb{Z}_{\geq 0}\to\mathbb{C}$ to the functional equation $$f(m^2+n^2)=\big(f(m)\big)^2+\big(f(n)\big)^2$$ for all $m,n\in\mathbb{Z}_{\geq 0}$ are

• $f(n)=0$ for all $n\in\mathbb{Z}_{\geq 0}$,
• $f(n)=\dfrac12\,g(n)$ for all $n\in\mathbb{Z}_{\geq 0}$ where $g:\mathbb{Z}_{\geq 0}\to \{-1,+1\}$ is such that $g(s)=1$ for all $s\in S$, and
• $f(n)=n\,g(n)$ for all $n\in\mathbb{Z}_{\geq 0}$ where $g:\mathbb{Z}_{\geq 0}\to \{-1,+1\}$ is such that $g(s)=1$ for all $s\in S$.