# $f(f(x)) = 1 + x^2$, then what is f(1)?

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.

• If one could read Chinese or use a translator, there's this same discussion on Zhihu: zhihu.com/question/340104755 – Edward H. Aug 14 at 6:09
• 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 at 6:23
• @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 $(1,5)\setminus\{2\}$ – Edward H. Aug 14 at 6:26
• 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 at 6:38
• – Feng Shao Aug 14 at 8:15