All real values of $x$ which satisfy the equation $\sqrt{1+\sqrt{1+\sqrt{1+x}}}=x$

$\bf{My\; Try::}$ Here $\sqrt{1+\sqrt{1+\sqrt{1+x}}} = x>0$

Now Let $f(x)=\sqrt{1+x}\;,$ Then equation convert into $f(f(f(x)))=x$

Now Here $f(x)=x$ be one function which satisfy above equation.

My question is how can we calculate other function which satisfy above functional equation.

Help required, Thanks

  • 1
    $\begingroup$ If You had $f(x)=x$, for some $x$, then $f^{\circ 3}(x)=x$ as well. So you can find one solution by solving $x=\sqrt{1+x}$. You are trying to solve a degree 8 equation, and finding the corresponding solutions will reduce it to a degree 4 equation. $\endgroup$
    – QTHalfTau
    Jun 22, 2016 at 5:24
  • $\begingroup$ This is a good question! $\endgroup$
    – Ghartal
    Jun 22, 2016 at 5:50
  • $\begingroup$ the title of the post should match the question in the post. $\endgroup$
    – miracle173
    Jun 22, 2016 at 6:34
  • 1
    $\begingroup$ There is no substantial relationship between finding $x$ which satisfies $f(f(f(x)))=x$ where $f(x)=\sqrt{1+x}$ and finding a function $f(x)$ which satisfies $f(f(f(x)))=x$ for all $x$ $\endgroup$
    – Henry
    Jun 22, 2016 at 7:15
  • 1
    $\begingroup$ @Ghartal: I don't think so. If you read the comments to the question and the answer you can see that the question is very unclear. Actually the answer consists of two answers because Jennifer is not sure what the question is. The OP ignores that this question in unclear and does not try to improve this. I think the best is to vote it down until the OP improves the question. $\endgroup$
    – miracle173
    Jun 22, 2016 at 7:56

3 Answers 3


With your remark $x=\sqrt{1+x}$ should lead to a solution.

$$x=\sqrt{1+x} \iff x^2=1+x \iff x^2-x-1=0 \iff x=\frac{1\pm\sqrt5}{2}$$

But $x=\frac{1+\sqrt5}{2}$ is the only positive solution. We verify that it is indeed solution by using long division :

$$\sqrt{1+\sqrt{1+\sqrt{1+x}}} = x\iff \sqrt{1+\sqrt{1+x}}=x^2-1 \iff \sqrt{1+x}=x^4-2x^2\\\iff x+1=x^8-4x^6+4x^4 \iff x^8-4x^6+4x^4-x-1=0$$


Let $g(x)=x^6+x^5-2x^4-x^3+x^2+1$, notice that $\forall x>0, g'(x)>0$, so $g$ is strictly increasing and $g(0)=1>0$, so $g(x)=0$ has no solution on $\mathbb{R}^{*+}$.

Finally $x=\frac{1+\sqrt5}{2}$ is the only solution to the problem.

Notes : The functional equation $\;(f \circ f \circ f)(x) = x,\forall x>0$ only has one continuous solution : $f(x)=x$.

By studying the domain, you have $f : \mathbb{R}^{*+} \rightarrow \mathbb{R}^{*+}$.

$\forall x \in \mathbb{R}^{*+}$ the image by $f$ of $f(f(x))$ is $x$ so $f$ is onto. Also if $f(a)=f(b)$, then $a=f(f(f(a)))=f(f(f(b)))=b$, so $f$ is injective. So $f$ is one to one. If $f$ is decreasing $f\circ f$ is increasing and $f\circ f\circ f$ is decrasing, but $id$ is increasing, it is absurd, so $f$ is strictly increasing.

Suppose $f(x)>x$, since $f$ is strictly increasing $f(f(x))>f(x)>x$, so $x=f(f(f(x))>f(f(x))>f(x)>x$, it is impossible. In the same way $f(x)<x$ is impossible, so $\forall x, f(x)=x$. So $f=id$.

  • 2
    $\begingroup$ Your reasoning is incorrect. In the link, $f$ satisfies the property for ALL $x$. $\endgroup$
    – Aravind
    Jun 22, 2016 at 5:36
  • $\begingroup$ @Aravind I posted an adapted proof $\endgroup$
    – Bérénice
    Jun 22, 2016 at 5:59
  • $\begingroup$ Ah! I see now. Here, we already know that $f$ is strictly increasing in $x > 0$, so we could also use that. $\endgroup$
    – Aravind
    Jun 22, 2016 at 6:18
  • $\begingroup$ @EricWofsey I have to justify why only $x=\sqrt{1+x}$ is the only expression which leads to solutions, I don't see why it should be removed, can you explain me please ? $\endgroup$
    – Bérénice
    Jun 22, 2016 at 6:27
  • $\begingroup$ The problem statement does not tell you that $f(f(f(x)))=x$ for all $x$, it just asks you to find the values of $x$ for which this is true. $\endgroup$ Jun 22, 2016 at 6:30

\begin{align}\sqrt{1+{\sqrt{1+\sqrt{1+x}}}}=x&\implies \sqrt{1+{\sqrt{1+\sqrt{1+x}}}}=x\\&\implies \left(1+{\sqrt{1+\sqrt{1+x}}}\right)=x^2\\&\implies\sqrt{1+\sqrt{1+x}}=x^2-1\\&\implies 1+\sqrt{1+x}=x^4-2x^2+1\\&\implies \sqrt{1+x}=x^4-2x^2\\&\implies 1+x=x^4(x^2-2)^2\\&\implies 1+x=x^4(x^4-4x^2+4)\\&\implies x^8-4x^6+4x^4-x-1=0\end{align}Now observe that , $$x^8-4x^6+4x^4-x-1=x^8\color{red}{-x^7}-x^6\color{red}{+x^7}-x^6\color{blue}{-x^5}-2x^6\color{blue}{+2x^5}+2x^4\color{blue}{-x^5}+x^4\color{green}{+x^3}+x^4\color{green}{-x^3}\color{violet}{-x^2}\color{violet}{+x^2}-x-1$$Consequently, $$x^8-4x^6+4x^4-x-1=0\implies (x^2-x-1)(x^6+x^5-2x^4-x^3+x^2+1)$$Now observe that the function $$g(x)=x^6+x^5-2x^4-x^3+x^2+1=\left(x^3-1\right)^2+\left(x-1\right)^2\,x^3+x^2 >0$$ for all $x\in \mathbb{R}^+$. So $(x^2-x-1)=0$ and hence $x=\dfrac{1+\sqrt{5}}{2}$ is the only positive solution.

  • 1
    $\begingroup$ How do you know $g(x)>0$ for all $x\in\mathbb{R}$? $\endgroup$ Jun 22, 2016 at 7:51
  • $\begingroup$ Isn't this answer already included in Jennifer's answer? $\endgroup$
    – miracle173
    Jun 22, 2016 at 8:06
  • $\begingroup$ @miracle173: Actually. I didn't notice the edit that was being made in Jennifer's answer as I was writing my answer. So, the answer to your question is yes. $\endgroup$
    – S. Das
    Jun 22, 2016 at 8:13
  • $\begingroup$ @EricWofsey: edited the post so one can see that g(x)>0 $\endgroup$
    – miracle173
    Jun 22, 2016 at 12:51

Let $1+x=y^2$, where $y>0$ and $1+y=z^2$, where $z>0$.

Hence, $1+z=x^2$, where $x>0$.

Thus, $x-y=(y-z)(y+z)$ and $y-z=(z-x)(z+x)$.

  1. Let $x>y$. Hence, $y>z$, which says that $z>x$. It's contradiction.

  2. Let $x<y$. Hence, $y<z$, which says that $z<x$. It's contradiction again.

Id est, $x=y=z$ and $x=\frac{1+\sqrt5}{2}$.


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