Values of $a$ for Which $|||x-1|-3|-a|=k$ Has 8 Distinct Real Roots Question
$|||x-1|-3|-a|=k$, $a \in \mathbb{N}$ has 8 distinct real roots for some $k$, then find the number of such values of $a$.
My Thought
Sorry I cannot show any work this time because I don't understand how to proceed. Please provide some guidance.
 A: Let $y=|||x-1|-3|-a|$ To have $y=k$ have 8 distinct roots , it is obvious that $y=0$ must have 4 distinct roots. 
Now check the graph of $||x-1|-3|$ .Only if $0\lt a\lt3$ does the previous condition satisfy. So $a=1$ OR $a=2$
A: Of course, we have to assume that $k \ge 0$.
\begin{align}
    |||x-1|-3|-a| &= k \\
      ||x-1|-3|-a &= \pm k \\
        ||x-1|-3| &= a \pm k \\
          |x-1|-3 &= \pm(a \pm k) \\
            |x-1| &= 3 \pm(a \pm k) \\
              x-1 &= \pm(3 \pm(a \pm k)) \\
                x &= 1 \pm(3 \pm(a \pm k)) \\
\end{align}
You get $x \in \left\{ \begin{matrix}
    1+3+a+k,\\
    1+3+a-k,\\
    1+3-a+k,\\
    1+3-a-k,\\
    1-3+a+k,\\
    1-3+a-k,\\
    1-3-a+k,\\
    1-3-a-k\\
\end{matrix} \right\}$
which simplifies to $x \in \left\{ \begin{matrix}
    4+a+k,\\
    4+a-k,\\
    4-a+k,\\
    4-a-k,\\
    -2+a+k,\\
    -2+a-k,\\
    -2-a+k,\\
    -2-a-k\\
\end{matrix} \right\}$
For unique roots, you need to find when any pair of solutions is equal and restrict $a$ accordingly.
Being a bit free and easy with the $\pm$s, $4\pm a\pm k = 4\pm a\pm k$ leads to $a \not \in \{k, -k\}$. The same goes for $-2\pm a\pm k = -2\pm a\pm k$. The $16$ equations $4\pm a\pm k = -2\pm a\pm k$ lead to 
$a \not \in \{k+3, k-3, -k+3, -k-3\}$. There will also not be eight unique solutions when $k \in \{0, 3, -3\}$.
