TITLE: Finding all solutions to the equation $|||||x|-1|-1|-1|-1|=0$
QUESTION [15 upvotes]: I was presented this question by a student I was tutoring:

Suppose $x \in \mathbb{R}$. Find all solutions of the equation $$|||||x|-1|-1|-1|-1|=0.$$ 

What I explained to the student:
Given $|||||x|-1|-1|-1|-1|=0$, one can drop the outermost absolute value bars, and place the plus/minus symbol on the right side, like this: $$||||x|-1|-1|-1|-1=\pm 0=0 \implies ||||x|-1|-1|-1|=1.$$
Repeating this process, $$|||x|-1|-1|-1=\pm 1 \implies |||x|-1|-1|=0,2.$$
And again, $$||x|-1|-1=0,2 \implies |||x|-1|=1,3.$$
And yet again, $$|x|-1=\pm1, \pm3 \implies |x|=0,2,4.$$
Finally, we arrive with $\boxed{x=-4,-2,0,2,4}$.
Was this approach okay? How did I do as far as finding the solutions go?

REPLY [7 votes]: Your calculations look solid to me. Since it seems like you understand the computational aspects of this type of problem, I'd just like to offer a geometric perspective on the problem. Here's a graph of the function $f(x)=|||||x|-1|-1|-1|-1|$ for $-7<x<7$:

The five points where the function meets the $x$-axis correspond to the five roots of the equation you calculated. Note the periodic saw-tooth nature of the function between $-4$ and $4$. Once you know that $x=0$ is a root and $x=2$ is a root, you can automatically infer from symmetry that $x=-2,x=4,x=-4$ are also roots. All that's left at this point is logically justifying that these are the only roots.