Equation with huge number of nested square roots:

Find all real roots of equation: $$ \underbrace{\sqrt{x+2\sqrt{x+2\sqrt{x+...+2\sqrt{x+2\sqrt{3x}}}}}}_\text{2018 roots}=x $$

I know that the answers are 0 and 3. What is common about those values is that each time, the square root is applied, the result is exactly $x$, therefore the number of iterations doesn't really matter.

To clarify that, declare the sequence of $a_i(x)$ as: $$ a_0(x)=x;\quad a_n(x)=\sqrt{x+2a_{n-1}(x)} $$ So, the equation of the problem can be written as $a_{2018}(x)=x$.

If it could be proven that for given $x$ the sign of $a_n-a_{n-1}$ is same for any $n$ (i.e. the sequence is monotonously increasing, decreasing or constant) then, since $a_{2018}(x)=a_{0}(x)$ the sequence must be constant, in particular $a_1(x)=a_0(x)$, ie. $\sqrt{3x}=x$, from which it follows that $x$ must be either $0$ or $3$. And indeed, it can be easily checked that the sequence is constant for those values.

To my disappointment, I can't find a way to prove that sign of $a_n-a_{n-1}$ is same. It may be really wrong.

Solution 1:

Hint: $\displaystyle\;\; a_n-a_{n-1}=\sqrt{x+2a_{n-1}}-\sqrt{x+2a_{n-2}}=\frac{2(a_{n-1}-a_{n-2})}{\sqrt{x+2a_{n-1}}+\sqrt{x+2a_{n-2}}}\,$, so the differences between consecutive terms have the same sign.

Solution 2:

Rather than comparing $a_n$ to $a_{n-1}$, it is much simpler to compare $a_n$ to $x$. Observe that if $0<x<3$ and $y\geq x$ then $$\sqrt{x+2y}\geq\sqrt{3x}>\sqrt{x^2}=x.$$ It follows that if $0<x<3$, then $a_n(x)>x$ for all $n>0$ (by induction on $n$). Similarly (just reverse all the inequalities), if $x>3$, then $a_n(x)<x$ for all $n>0$. So the only way we can ever have $a_n(x)=x$ for any $n>0$ is if $x=3$ or $x=0$.