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I get $f(f(a)) = a^2 + 1 = f(f(-a))$, and so $f(a)^2 + 1 = f(a^2 + 1) = f(-a)^2 + 1$, so $f(a) = f(-a)$ or $f(a) = -f(-a)$, but then I donot know what to do next. Thanks for any help.

Jean-Claude Arbaut
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Ziqin He
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    If one could read Chinese or use a translator, there's this same discussion on Zhihu: https://www.zhihu.com/question/340104755 –  Aug 14 '19 at 06:09
  • Above link shows how to construct a solution of the given functional equation. In fact there is a construction in English also. But I don't see uniqueness there. Is it clear from that post that $f(1)$ is uniquely determined? – Kavi Rama Murthy Aug 14 '19 at 06:23
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    @KaviRamaMurthy Ah no, the second highest voted answer in the same thread showed that if we drop analyticity of $f$ then it's anywhere in $(1,5)\setminus{2}$ –  Aug 14 '19 at 06:26
  • OP should provide some context. If $f(1)$ cannot be determined uniquely by the given equation I don't think this question is appropriate. – Kavi Rama Murthy Aug 14 '19 at 06:38
  • See https://www.jstor.org/stable/2321556?seq=1#page_scan_tab_contents. – Feng Aug 14 '19 at 08:15
  • What are the domain and codomain of $f$ supposed to be? $\mathbb{R}$? – Eric Wofsey Apr 15 '20 at 02:54
  • See https://math.stackexchange.com/questions/1612308/if-ffx-x2-1-then-f6 and https://math.stackexchange.com/questions/3633/on-the-functional-square-root-of-x21 – Gerry Myerson Apr 15 '20 at 04:27

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A piece of information is still required

$f(f(x)) = 1+x^2$

Let's say $y = f(x)$ is the function, then $y = g(x)$ is the inverse function

$f(x) = g(1+x^2)$, $f(1) = g(2)$

$f(f(x)) = 1+x^2 = f(f(-x))$ $f(f(x)) = f(f(-x))$ Therefore $f(x) = f(-x)$ $f(x)$ is a even function

$f(f(x)) = 1+x^2$, say $x → g(x)$

$f(x) = 1+g(x)^2$, $f(1) = 1+g(1)^2$

$1+g(x)^2 = g(1+x^2)$

Say $x → f(x)$ $1+x^2 = g(1+f(x)^2)$, $f(1+x^2) = 1+f(x)^2$

$f(x) = $

$1+f(x)^2 = f(1+x^2)$

I can't find $f(1)$ unless I knew $f(n)$

$f(1+x^2) = 1+f(x)^2$, Remember $f(x) = 1+g(x)^2$

$f(1+x^2) = 1+(g(x)^2+1)^2$, $f(1+x^2) = g(x)^4+2*g(x)^2+2$

Recall $f(x) = g(1+x^2)$, say $1+x^2→x$ $g(x) = f(\sqrt(x-1))$

$f(1+x^2) = f(\sqrt(x-1))^4+2*f(\sqrt(x-1))^2+2$

Ian
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Aderinsola Joshua
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