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

Question

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

Answer 1

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 :

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

$$x^8-4x^6+4x^4-x-1=(x^2-x-1)(x^6+x^5-2x^4-x^3+x^2+1)$$

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.

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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 $f=id$.

Answer 2

\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.

Answer 3

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 $x=\frac{1+\sqrt5}{2}$.