Finding all solutions to the equation $|||||x|-1|-1|-1|-1|=0$

Question

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?

Answer 1

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.

Answer 2

Let $f(x) = ||x|-1|$. You have $f(x) = c$ iff $x=\pm(c+1)$ or $x=\pm (1-c)$.

The problem is to solve $f(f(f(f(x)))) = 0$.

This gives the possible values of $f(f(f(x)))$ as $\pm 1$.

This gives the possible values of $f(f(x))$ as $0, \pm 2$.

This gives the possible values of $f(x)$ as $\pm 1, \pm 3$.

Finally, this gives the possible values of $x$ as $0, \pm 2, \pm 4$.

The general pattern follows from this...