10
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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.

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    $\begingroup$ If one could read Chinese or use a translator, there's this same discussion on Zhihu: zhihu.com/question/340104755 $\endgroup$ – Edward H. Aug 14 at 6:09
  • $\begingroup$ 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? $\endgroup$ – Kavi Rama Murthy Aug 14 at 6:23
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    $\begingroup$ @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 $(1,5)\setminus\{2\}$ $\endgroup$ – Edward H. Aug 14 at 6:26
  • $\begingroup$ OP should provide some context. If $f(1)$ cannot be determined uniquely by the given equation I don't think this question is appropriate. $\endgroup$ – Kavi Rama Murthy Aug 14 at 6:38
  • $\begingroup$ See jstor.org/stable/2321556?seq=1#page_scan_tab_contents. $\endgroup$ – Feng Shao Aug 14 at 8:15

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