# functional square root problem - Solution with "Chain Solution". - 2 example problems: $f(f(x))=x^2-2$, $f(f(x))-x^2+x+3=0$ + 1 more problem

Is there a function / Are there such functions $$f: \mathbb{R} \to \mathbb{R}$$ which satisfies $$f(f(x))=x^2-2$$?

My Solution:

\begin{align} &\text{Let's look for t_2 which satisfies } f^2(t_2)=t_2. \\ &\Rightarrow t_2=2, -1. \\ \ \\ &\text{Now, let's look for t_4 which satisfies } f^4(t_4)=t_4. \\ &\Rightarrow f^4(t_4)=f^2(t_4^2-2) = t_4^4-4t_4^2+2. \\ &\text{let set }T_2 = \{t_2\}, T_4=\{t_4\}. \\ &\Rightarrow T_2 \subset T_4. \\ &n(T_4)=4, n(T_2)=2. \Rightarrow \text{let } T_4=\{2, -1, \alpha, \beta \}. (\alpha, \beta \not\in T_2.) \\ \ \\ &\text{I will define the 2chain, -1chain, \alphachain and \betachain:} \\ \ \\ &\text{2chain}: \begin{bmatrix} 2 & \rightarrow & f(2) \\ \uparrow & \fbox{2_\text{cyc}} & \downarrow \\ f(2) &\leftarrow & 2 \end{bmatrix} \\ \ \\ -&\text{1chain}: \begin{bmatrix} -1 & \rightarrow & f(-1) \\ \uparrow & \fbox{2_\text{cyc}} & \downarrow \\ f(-1) & \leftarrow & -1 \end{bmatrix} \\ \ \\ &\text{\alphachain}: \begin{bmatrix} \alpha & \rightarrow & f(\alpha) \\ \uparrow & \fbox{4_\text{cyc}} & \downarrow \\ f^3(\alpha) & \leftarrow & f^2(\alpha) \end{bmatrix} \\ \ \\ &\text{\betachain}: \begin{bmatrix} \beta & \rightarrow & f(\beta) \\ \uparrow & \fbox{4_\text{cyc}} & \downarrow \\ f^3(\beta) & \leftarrow & f^2(\beta) \end{bmatrix} \\ \ \\ &\text{Let's look at \alphachain.} \\ \ \\ &\text{if } f(\alpha)=\alpha \Rightarrow f^2(\alpha)=\alpha, \text{Contradiction.} \\ &\text{if } f(\alpha)=2 \Rightarrow f^3(\alpha)=2, f^2(\alpha)=\alpha, \text{Contradiction.} \\ &\text{if } f(\alpha)=-1 \Rightarrow f^3(\alpha)=-1, f^2(\alpha)=\alpha, \text{Contradiction.} \\ &\text{if } f(\alpha)=\beta \Rightarrow \text{Let's look at \betachain together.} \\ \ \\ &\text{if } f(\beta)=\beta \Rightarrow f^2(\beta)=\beta, \text{Contradiction.} \\ &\text{if } f(\beta)=2 \Rightarrow f^3(\beta)=2, f^2(\beta)=\beta, \text{Contradiction.} \\ &\text{if } f(\beta)=-1 \Rightarrow f^3(\beta)=-1, f^2(\beta)=\beta, \text{Contradiction.} \\ &\text{if } f(\beta)=\alpha \Rightarrow \text{Going back to the \alphachain, } f^2(\alpha)=\alpha, \text{Contradiction.} \\ \ \\ &\therefore \not\exists f: \mathbb{R} \to \mathbb{R} \text{ s.t. } f(f(x))=x^2-2. \end{align}

P.S.

1. You can prove this with $$f^3(\alpha)$$, too. I recommend trying this!

Prove that there isn't such function $$f: \mathbb{R} \to \mathbb{R}$$ satisfying $$f(f(x))-x^2+x+3=0$$.

My Solution:

\begin{align} &\text{Let's take a look at t_2 which satisfies f^2(t_2)=t_2.} \\ &\Rightarrow t_2^2-2t_2-3=0, t_2=-1, 3. \\ \ \\ &\text{Now, let]s take a look at t_4 which satisfies f^4(t_4)=t_4.} \\ &\Rightarrow (t_4^2-t_4-3)^2-(t_4^2-t_4-3)-3=t_4. \\ &\therefore t_4 \text{ has 4 solution.}\\ \ \\ &\text{let set T_2=\{t_2\}, T_4=\{t_4\}.} \\ &T_2 \subset T_4, \ n(T_2)=2, \ n(T_4)=4. \Rightarrow T_2=\{-1, 3\}, T_4=\{-1, 3, \alpha, \beta \} (\alpha, \beta \neq -1, 3.) \\ \ \\ &\text{Now, I will define -1chain, 3chain, \alphachain and \betachain:} \\ \ \\ -&\text{1chain}: \begin{bmatrix} -1 & \rightarrow & f(-1) \\ \uparrow & \fbox{2_\text{cyc}} & \downarrow \\ f(-1) & \leftarrow & -1 \end{bmatrix} \\ \ \\ &\text{3chain}: \begin{bmatrix} 3 & \rightarrow & f(3) \\ \uparrow & \fbox{2_\text{cyc}} & \downarrow \\ f(3) & \leftarrow & 3 \end{bmatrix} \\ \ \\ &\text{\alphachain}: \begin{bmatrix} \alpha & \rightarrow & f(\alpha) \\ \uparrow & \fbox{4_\text{cyc}} & \downarrow \\ f^3(\alpha) & \leftarrow & f^2(\alpha) \end{bmatrix} \\ \ \\ &\text{\betachain}: \begin{bmatrix} \beta & \rightarrow & f(\beta) \\ \uparrow & \fbox{4_\text{cyc}} & \downarrow \\ f^3(\beta) & \leftarrow & f^2(\beta) \end{bmatrix} \\ \ \\ &\text{Let's look at \alphachain.} \\ \ \\ &\text{if } f(\alpha)=\alpha \Rightarrow f^2(\alpha)=\alpha, \text{Contradiction.} \\ &\text{if } f(\alpha)=-1 \Rightarrow f^3(\alpha)=-1, f^2(\alpha)=\alpha, \text{Contradiction.} \\ &\text{if } f(\alpha)=3 \Rightarrow f^3(\alpha)=3, f^2(\alpha)=\alpha, \text{Contradiction.} \\ &\text{if } f(\alpha)=\beta \Rightarrow \text{Let's look at \betachain together.} \\ \ \\ &\text{if } f(\beta)=\beta \Rightarrow f^2(\beta)=\beta, \text{Contradiction.} \\ &\text{if } f(\beta)=-1 \Rightarrow f^3(\beta)=-1, f^2(\beta)=\beta, \text{Contradiction.} \\ &\text{if } f(\beta)=3 \Rightarrow f^3(\beta)=3, f^2(\beta)=\beta, \text{Contradiction.} \\ &\text{if } f(\beta)=\alpha \Rightarrow \text{Going back to the \alphachain, } f^2(\alpha)=\alpha, \text{Contradiction.} \\ \ \\ &\therefore \not\exists f: \mathbb{R} \to \mathbb{R} \text{ s.t. } f(f(x))-x^2+x+3=0. \end{align}

P.S. If you want, you can try solving this one with a chain solution:

$$f(f(x))=x^2-14x+56.$$

\begin{align} &\text{Let's look for t_2 which satisfies f^2(t_2)=t_2.} \\ &\Rightarrow t_2=7, 8. \\ \ \\ &\text{Now, let's look for t_4 which satisfies f^4(t_4)=t_4.} \\ &\Rightarrow f^4(t_4)=f^2(t_4^2-14t_4+56)=(t_4^2-14t_4+56)^2-14(t_4^2-14t_4+56)+56. \\ & \therefore \text{t_4 has 4 solution.} \\ \ \\ &\text{let set T_2=\{t_2\}, T_4=\{t_4\}.} \\ &\Rightarrow n(T_2)=2, n(T_4)=4, T_2 \subset T_4. \Rightarrow \text{let }T_4-T_2=\{\alpha, \beta \}. \\ & \Rightarrow T_2=\{7, 8\}, T_4=\{7, 8, \alpha, \beta \}. \\ \ \\ &\text{I will define the 7chain, 8chain, \alphachain and \betachain:}\\ \ \\ & \text{7chain}: \begin{bmatrix} 7 & \rightarrow & f(7) \\ \uparrow & \fbox{2_\text{cyc}} & \downarrow \\ f(7) &\leftarrow & 7 \end{bmatrix} \\ \ \\ &\text{8chain}: \begin{bmatrix} 8 & \rightarrow & f(8) \\ \uparrow & \fbox{2_\text{cyc}} & \downarrow \\ f(8) &\leftarrow & 8 \end{bmatrix} \\ \ \\ &\text{\alphachain}: \begin{bmatrix} \alpha & \rightarrow & f(\alpha) \\ \uparrow & \fbox{4_\text{cyc}} & \downarrow \\ f^3(\alpha) & \leftarrow & f^2(\alpha) \end{bmatrix} \\ \ \\ &\text{\betachain}: \begin{bmatrix} \beta & \rightarrow & f(\beta) \\ \uparrow & \fbox{4_\text{cyc}} & \downarrow \\ f^3(\beta) & \leftarrow & f^2(\beta) \end{bmatrix} \\ \ \\ &\text{if } f(\alpha)=\alpha \Rightarrow f^2(\alpha)=\alpha, \text{Contradiction.} \\&\text{if } f(\alpha)=7 \Rightarrow f^3(\alpha)=7,f^2(\alpha)=\alpha, \text{Contradiction.} \\&\text{if } f(\alpha)=8 \Rightarrow f^3(\alpha)=8, f^2(\alpha)=\alpha, \text{Contradiction.} \\ &\text{if } f(\alpha)=\beta \Rightarrow \text{Let's look at \betachain together.} \\ \ \\ &\text{if } f(\beta)=\beta \Rightarrow f^2(\beta)=\beta, \text{Contradiction.} \\ &\text{if } f(\beta)=7 \Rightarrow f^3(\beta)=7, f^2(\beta)=\beta, \text{Contradiction.} \\ &\text{if } f(\beta)=8 \Rightarrow f^3(\beta)=8, f^2(\beta)=\beta, \text{Contradiction.} \\ &\text{if } f(\beta)=\alpha \Rightarrow \text{Going back to the \alphachain, } f^2(\alpha)=\alpha, \text{Contradiction.} \\ \ \\ \end{align}
• Does this answer your question? How to solve the functional equation $f(f(x))=ax^2+bx+c$ Nov 17, 2021 at 13:01
• I agree with a comment on the previous question that the "chain" notation could be better. $x\to f(x)\to f^2(x)\to f^3(x)\to x$, for example. The notation you have tried looks like the notation for a matrix with the specified elements in the four corners, where the "$\ldots$" indicate some number of rows or columns that are to be filled in somehow. And of course your method has nothing to do with such a matrix. Nov 17, 2021 at 13:52