# Functional equation: composition to get quadratic

Consider the following functional equation: $$f(f(x))=x^2+x-7\quad\quad\forall\; x\in\mathbb{R}.$$ Does there exist a function $$f:\mathbb{R}\to\mathbb{R}$$ satisfying this, or not?

• This puzzle may be a bit too hard. If it goes unsolved for a while, I'll add a hint. Feb 5, 2021 at 14:09
• rot13(Guvf cqs fubhyq or bs uryc: uggc://lnebfyniio.pbz/cncref/evpr-jura.cqs)
– user
Feb 5, 2021 at 14:33
• @user Oh interesting! That solves the problem for sure, but it can be done (at least in this case) without using all of that complex machinery. Feb 5, 2021 at 14:36
• If $x = 0$, we have the property $f(f(0)) = -7$. Wonder if that helps? Feb 5, 2021 at 15:13
• Ah, but while the main theorem proved in the article is about functions C -> C, the last section comments briefly on the R->R situation and there's a theorem there that resolves this problem. Feb 5, 2021 at 17:41

Let's look at fixed points. And let's write $$f^2(x)$$ for $$f(f(x))$$.
$$f^2(x)=x$$ has two fixed points: $$\pm\sqrt 7$$. These are also fixed points of $$f^4(x)$$, together with $$-1\pm\sqrt 6$$, which are swapped by $$f^2(x)$$.
As $$f$$ maps any fixed point of $$f^n$$ to a fixed point of $$f^n$$, $$f$$ restricted to the four points $$\pm\sqrt 7,-1\pm\sqrt 6$$ is a permutation. $$f^2$$ being a square must be even but we have just seen that $$f^2$$ is an elementary permutation, hence odd. Contradiction.