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Remark. Dear voters, this question is not a duplicate of the following old question. Please refrain from closing it for being a duplicate.

QUESTION: Find all functions $f:\Bbb{N}→\Bbb{N}$ which satisfy $$f(m^2+n^2)=f(m)^2+f(n)^2\,,\forall\space m,n\in\Bbb{N}\,.$$ Here, $\mathbb{N}=\{1,2,3,\ldots\}$.


MY APPROACH: Set $m=n$.. we obtain- $$f(2n^2)=2f(n)^2$$ Studying this equation I found out that $f(x)=\sqrt{\frac{x}2}$ satisfies the condition. But I could not think any further. If my claim is true, how do I prove it? And how do I make sure that there aren't any other functions satisfying the property?

Thank you for your help in advance :)

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    $\begingroup$ Does your $N$ contains $0$? $\endgroup$ Jul 4, 2020 at 6:54
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    $\begingroup$ Then why is the second condition even given.Seems weird. $\endgroup$ Jul 4, 2020 at 6:56
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    $\begingroup$ @Yes it's me yes.. I too didn't get why was the second condition is given.. anyway, do you have any idea how to tackle this? $\endgroup$ Jul 4, 2020 at 6:59
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    $\begingroup$ Please always state whether $\mathbb{N}=\{1,2,3,\ldots\}$ or $\mathbb{N}=\{0,1,2,3,\ldots\}$. People use different conventions for $\mathbb{N}$. Your clarification of what $\mathbb{N}$ is is necessary. $\endgroup$ Jul 4, 2020 at 7:42
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    $\begingroup$ I think he means $f:\mathbb N_0\to\mathbb N_0$ as this way, the problem makes more sense. $\endgroup$
    – Anand
    Jul 4, 2020 at 8:17

2 Answers 2

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The only such function is $f(n)=n$.

The hard part is to show $f(1)=1$. After that, one just follows the answer to Find all $f$ such that $f\left(m^{2}+n^{2}\right)=f(m)^{2}+f(n)^{2},$ plus some small values of $f(n)=n$ that will be proved below.

Let $f(1)=a$.

The given formula applies directly to any number that is the sum of two squares. Thus \begin{align*} f(2)&=f(1+1)=2a^2\\ f(5)&=f(4+1)=(2a^2)^2+a^2=4a^4+a^2\\ f(8)&=f(4+4)=8a^4\\ f(10)&=f(9+1)=f(3)^2+a^2\\ f(13)&=f(9+4)=f(3)^2+4a^4 \end{align*}

However, some numbers are the sum of two squares in more than one way, and this is the key to the proof.

\begin{align} 7^2+1&=5^2+5^2,&f(7)^2&=2f(5)^2-a^2\\ 2^2+11^2&=5^2+10^2,&f(11)^2&=f(10)^2+f(5)^2-f(2)^2\\ 11^2+7^2&=13^2+1,&f(11)^2&=f(13)^2-f(7)^2+a^2\\ \hline 5^2+14^2&=10^2+11^2,&f(14)^2&=f(10)^2+f(11)^2-f(5)^2\\ 6^2+13^2&=3^2+14^2,&f(6)^2&=f(14)^2+f(3)^2-f(13)^2\\ \end{align}

The first three equations involve the variables $f(3)^2$, $f(7)^2$, $f(11)^2$. Eliminating $f(11)^2$ and $f(7)^2$ gives $$a^2 (4 a^2-1)f(3)^2 = a^2 (2a^2+1)^2(4a^2-1)$$ so $f(3)=2a^2+1$.

We can now use the last two equations to find $f(14)^2$, then $$f(6)^2=2(1+8a^2+17a^4+8a^6-16a^8)$$ This forces $a=1$ since the polynomial takes negative values for $a>1$.

It then follows that $f(n)=n$ for small values of $n$, by substitution, and hence by induction.

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Here we shall find all functions $f:\mathbb{Z}_{>0}\to\mathbb{C}$ such that $$f(m^2+n^2)=\big(f(m)\big)^2+\big(f(n)\big)^2$$ for all positive integers $m$ and $n$. Let $S$ denote the subset of $\mathbb{Z}_{>0}$ consisting of every integer that can be written as a sum of two perfect squares of positive integers. We shall prove that all solutions are given by

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

As in Chrystomath's answer, if we let $a:=f(1)$, then $$a^2(4a^2-1)\,\big(f(3)\big)^2=a^2(4a^2-1)^2\,(2a^2+1)^2\,.$$ This shows that $a=0$, $a=\pm\dfrac12$, or $f(3)=\pm(2a^2+1)$.

Case I: $a=0$. Then, Chrystomath's answer shows that $f(2)=f(5)=f(7)=f(8)=0$. Let $b:=f(3)$. Then, Chrystomath's answer tells us that $f(10)=b^2$, $f(11)=\pm b^2$, $f(13)=b^2$, $f(14)=\pm\sqrt{2}b^2$, and $f(6)=\pm b\sqrt{b^2+1}$. From $6^2+7^2=85=2^2+9^2$, we see that $$f(9)=\pm\sqrt{\big(f(6)\big)^2+\big(f(7)\big)^2-\big(f(2)\big)^2}=\pm b\sqrt{b^2+1}\,.$$ Now, $4^2+7^2=65=1^2+8^2$ implies that $$f(4)=\pm\sqrt{\big(f(1)\big)^2+\big(f(8)\big)^2-\big(f(7)\big)^2}=0\,.$$ Consequently, $$f(17)=f(4^2+1^2)=\big(f(4)\big)^2+\big(f(1)\big)^2=0\,.$$ Now, $1^2+12^2=145=8^2+9^2$ and $5^2+15^2=250=9^2+13^2$, we conclude as before that $$f(12)=\pm b\sqrt{b^2+1}\text{ and }f(15)=\pm b\sqrt{2b^2+1}\,.$$ Using $1^2+17^2=290=11^2+13^2$, we obtain $$2b^4=\big(f(11)\big)^2+\big(f(13)\big)^2=\big(f(1)\big)^2+\big(f(17)\big)^2=0\,.$$ Therefore, $b=0$. It follows that $f(n)=0$ for $n=1,2,3,\ldots,15$. Following the induction step from this link, we conclude that $f(n)=0$ for every positive integer $n$.

Case II: $a=\pm\dfrac12$. Then, Chrystomath's answer shows that $$f(2)=f(5)=f(8)=\dfrac12\text{ and }f(7)=\pm\dfrac12\,.$$ Let $b:=f(3)$. Then, as in the previous case, $$f(10)=f(13)=b^2+\dfrac14\text{ and }f(11)=\pm\left(b^2+\dfrac14\right)\,.$$ Furthermore, $$f(14)=\pm\sqrt{2b^4+b^2-\frac18}\,.$$ and $$f(6)=\pm\sqrt{b^4+\frac{3}{2}b^2-\frac{3}{16}}\,.$$ Now, $4^2+7^2=65=1^2+8^2$ implies that $$f(4)=\pm\sqrt{\big(f(1)\big)^2+\big(f(8)\big)^2-\big(f(7)\big)^2}=\pm\frac12\,.$$ Consequently, $$f(17)=f(4^2+1^2)=\big(f(4)\big)^2+\big(f(1)\big)^2=\dfrac12.\,.$$ Using $1^2+17^2=290=11^2+13^2$, we obtain $$2\left(b^2+\dfrac14\right)^2=\big(f(11)\big)^2+\big(f(13)\big)^2=\big(f(1)\big)^2+\big(f(17)\big)^2=\dfrac12\,.\,.$$ Therefore, $b=\pm\dfrac12$ or $b=\pm\dfrac{\sqrt{3}\text{i}}{2}$. Using $4^2+17^2=305=7^2+16^2$, we conclude that $f(16)=\pm\dfrac12$. From $6^2+7^2=85=2^2+9^2$, we obtain $$f(9)=\pm\sqrt{b^4+\frac{3}{2}b^2-\frac{3}{16}}\,.$$ Using $1^2+12^2=145=8^2+9^2$, we conclude that $$f(12)=\pm\sqrt{b^4+\frac{3}{2}b^2-\frac{3}{16}}\,.$$ From $5^2+15^2=250=9^2+13^2$, we get $$f(15)=\pm\sqrt{2b^4+2b^2-\dfrac38}\,.$$

  • If $b=\pm\dfrac12$, then we can see that $f(n)=\pm\dfrac12$ for $n=1,2,3,\ldots,17$. By induction, we obtain $f(n)=\pm\dfrac12$ for all $n\in\mathbb{Z}_{>0}$. Furthermore, $f(n)=\dfrac12$ if $n$ can be written as a sum of two perfect squares of positive integers.

  • If $b=\pm\dfrac{\sqrt{3}\text{i}}{2}$, then that $$f(18)=f(3^2+3^2)=\big(f(3)\big)^2+\big(f(3)\big)^2=-\frac{3}{2}\,.$$ Now, $1^2+18^2=325=6^2+17^2$. This gives $$\big(f(1)\big)^2+\big(f(18)\big)^2=\big(f(6)\big)^2+\big(f(17)\big)^2$$ or $$\frac14+\frac{9}{4}=-\frac34+\frac14\,,$$ which is a contradiction. Therefore, this subcase leads to no solutions.

Case III: $a\notin \left\{0,\pm\dfrac12\right\}$. Then, $f(3)=\pm(2a^2+1)$. As in Chrystomath's answer, we can find $$f(2)=2a^2\,,\,\,f(5)=a^2(4a^2+1)\,,\,\,f(8)=8a^4\,,$$ $$f(10)=4a^4+5a^2+1\,,f(13)=8a^4+4a^2+1\,,$$ $$f(7)=\pm a\sqrt{32a^6+16a^4+2a^2-1}\,,$$ $$f(11)=\pm \sqrt{32a^8+48a^6+30a^4+10a^2+1}\,,$$ $$f(14)=\pm\sqrt{2(16a^8+40a^6+31a^4+10a^2+1)}\,,$$ $$f(6)=\pm\sqrt{2(-16a^8+8a^6+17a^4+8a^2+1)}\,.$$ Using $4^2+7^2=65=1^2+8^2$, we see that $$f(4)=\pm a\sqrt{2(16a^6-8a^4-a^2+1)}\,.$$ This gives $$\begin{align}f(17)&=f(1^2+4^2)=\big(f(1)\big)^2+\big(f(4)\big)^2\\&=a^2(32a^6-16a^4-2a^2+3)\,.\end{align}$$ Because $1^2+17^2=290=11^2+13^2$, we obtain $$(a-1)(a+1)(4a^2+1)^2\left(64a^{10}-32a^8+20a^6+16a^4+3a^2+2\right)=0\,.\tag{*}$$ From $6^2+7^2=85=2^2+9^2$, we have $$f(9)=\pm\sqrt{32a^6+32a^4+15a^2+2}\,.$$ Using $1^2+12^2=145=8^2+9^2$, we obtain $$f(12)=\pm\sqrt{64a^8+32a^6+32a^4+14a^2+2}\,.$$ From $5^2+15^2=250=9^2+13^2$, we get $$f(15)=\pm\sqrt{48a^8+88a^6+63a^4+23a^2+3}\,.$$

Now, we know $f(n)$ for $n=1,2,3,\ldots,15$. We can use the induction step from this link to determine $f(n)$ for $n=16,17,18,\ldots,26$, and obtain the following list. $$\begin{array}{|c|c|} \hline n&\big(f(n)\big)^2\\ \hline 1& a^2\\ 2& 4a^4\\ 3& 4a^4 + 4a^2 + 1\\ 4& 32a^8 - 16a^6 - 2a^4 + 2a^2\\ 5& 16a^8 + 8a^6 + a^4\\ 6& -32a^8 + 16a^6 + 34a^4 + 16a^2 + 2\\ 7& 32a^8 + 16a^6 + 2a^4 - a^2\\ 8& 64a^8\\ 9& 32a^6 + 32a^4 + 15a^2 + 2\\ 10& 16a^8 + 40a^6 + 33a^4 + 10a^2 + 1\\ 11& 32a^8 + 48a^6 + 30a^4 + 10a^2 + 1\\ 12& 64a^8 + 32a^6 + 32a^4 + 14a^2 + 2\\ 13& 64a^8 + 64a^6 + 32a^4 + 8a^2 + 1\\ 14& 32a^8 + 80a^6 + 62a^4 + 20a^2 + 2\\ 15& 48a^8 + 88a^6 + 63a^4 + 23a^2 + 3\\ 16& 96a^8 + 80a^6 + 58a^4 + 20a^2 + 2\\ 17& 96a^8 + 112a^6 + 62a^4 + 17a^2 + 2\\ 18& 64a^8 + 128a^6 + 96a^4 + 32a^2 + 4\\ 19& 96a^8 + 144a^6 + 90a^4 + 28a^2 + 3\\ 20& 144a^8 + 136a^6 + 89a^4 + 28a^2 + 3\\ 21& 96a^8 + 176a^6 + 122a^4 + 42a^2 + 5\\ 22& 128a^8 + 192a^6 + 124a^4 + 36a^2 + 4\\ 23& 160a^8 + 208a^6 + 122a^4 + 35a^2 + 4\\ 24& 160a^8 + 208a^6 + 150a^4 + 52a^2 + 6\\ 25& 176a^8 + 248a^6 + 151a^4 + 45a^2 + 5\\ 26& 160a^8 + 272a^6 + 182a^4 + 56a^2 + 6\\ 27& 192a^8 + 288a^6 + 184a^4 + 58a^2 + 7\\ 28& 256a^8 + 288a^6 + 180a^4 + 54a^2 + 6\\ 29& 224a^8 + 336a^6 + 210a^4 + 64a^2 + 7\\ 30& 208a^8 + 360a^6 + 245a^4 + 78a^2 + 9\\ \hline \end{array}$$

In particular, this shows that $$\big(f(26)\big)^2=160a^8+272a^6+182a^4+56a^2+6\,.$$ However, $$f(26)=f(1^2+5^2)=\big(f(1)\big)^2+\big(f(5)\big)^2\,,$$ so $$f(26)=a^2+(4a^4+a^2)^2=16a^8+8a^6+a^4+a^2\,.$$ This shows that $$\begin{align}(a-1)(a+1)(2a^2-a+1)&(2a^2+a+1)(4a^2+3)\\&(16a^8+8a^6+13a^4+12a^2+2)=0\,.\tag{#}\end{align}$$ From (*) and (#), we conclude that $a=\pm1$. This means $f(n)=\pm n$ for all positive integers $n$. Furthermore, $f(n)=n$ when $n$ can be written as a sum of two perfect squares of positive integers.

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