Equation involving recursive n:th-roots.

Question

Inspired by this question, replacing square roots with $n$-roots:

$$\sqrt[n]{X+\sqrt[n]{X+\sqrt[n]{X+ \dots}}} =\sqrt[n]{X\sqrt[n]{X\sqrt[n]{X \dots}}}$$ Then find the $X$ value?

Answer 1

This basically has the same solution: Solve $a=\sqrt[n]{X+a}$ for $a$. This gives you $a^n=a+X$, or $a^n-a=X$.

Now solve $b=\sqrt[n]{Xb}$ for $b$, this gives $b^n=Xb$. Now if we have $a=b$, then we have the following system of equations.

$$a^n-a=X$$

$$a^n=Xa$$

This gives $Xa-a=X$, and hence $X(a-1)=a$, thus $X=\frac{a}{a-1}$.

Now we have $a^n=\frac{a^2}{a-1}$, and hence $a^{n+1}-a^n-a^2=0$. Hence $a=0$ or $a^{n-1}-a^{n-3}-1=0$. The latter equation has no nice solutions unfortunately, but you can approximate them.