On solving $\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+x}}}}=\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+x}}}}$

How do we show that there is only one solution to,$$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+x}}}}=\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+x}}}}$$

I guess it is only $x=2$. Please help.

Solution 1:

Let $\;f(x) = \sqrt{2+x}\;$ and $\;g(x) = \sqrt[3]{6+x}$, they are strictly increasing function in $x$ when $x \ge -2$. Since $(x+2)^3 - (x+6)^2 = (x-2)(x^2 + 7x + 14)$ and $x^2 + 7x + 14 > 0$ for all $x$, we have $$\begin{cases} f(x) > g(x) > 2,& x > 2\\f(x) = g(x) = 2,& x = 2\\f(x) < g(x) < 2, & x <2\end{cases}$$ So for any $x > 2$, we have $$\begin{align} & f(x) > g(x) > 2\\ \implies & f(f(x)) > f(g(x)) > g(g(x)) > 2\\ \implies & f(f(f(x))) > f(g(g(x)) > g(g(g(x)) > 2\\ \implies & f(f(f(f(x))) > f(g(g(g(x))) > g(g(g(g(x)))) > 2\\ \implies & f(f(f(f(x))) \ne g(g(g(g(x)))) \end{align}$$ Please note that in above deductions, we are using following reasoning repeatedly. $$\underbrace{g\circ\cdots\circ g(x)}_{k \text{ terms}} > 2 \implies f(\underbrace{g\circ\cdots\circ g(x)}_{k \text{ terms}}) > \underbrace{g\circ\cdots\circ g(x)}_{k+1 \text{ terms}} > 2.$$

Similar logic shows that $f(f(f(f(x)))) \ne g(g(g(g(x))))$ for $x < 2$. As a result, $x = 2$ is the only solution for the equation $f(f(f(f(x)))) = g(g(g(g(x))))$.

Solution 2:

A proof by induction. Let: $$f_n(x) = \sqrt[3]{6+\sqrt[3]{6+\ldots+\sqrt[3]{6+x}}},\ g_n(x) = \sqrt{2+\sqrt{2+\ldots+\sqrt{2+x}}}$$ With $n$ terms. Then for $n=1$ you can easily solve the cubic equation to show that $f_1=g_1 $ only at $x=2$ (over the reals).

Now assume our claim is true for $n$, i.e. that $f_n(x)=g_n(x)$ iff for $x=2$. Then for $n+1$, raise to the sixth power to get that: $$(6+f_n(x))^2=(2+g_n(x))^3$$ Clearly, this equality is true for $x=2$ since $f_n(2)=g_n(2)$. Now, if our claim is false and this equality holds for some $x_0\neq 2$, then: $$g_n(x_0) = (6+f_n(x_0))^{2/3}-2$$ But since $dg_n/df_n < 1$ for all $x>0$, by the mean value theorem we have a contradiction. $$$$ Thus we have proven that for any number of $n$ the only (real) solution of $f_n(x)=g_n(x)$ is $x=2$.