How can I calculate real values of $x$ in $x \lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor = 88$, where $\lfloor x\rfloor$ is the floor function?
My attempt:
Let $\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor = k\in \mathbb{Z}$. Then
$$ k \leq x\lfloor x\lfloor x\rfloor\rfloor<k+1 $$
and our equation becomes
$$ x\cdot k = 88 \implies x=\frac{88}{k} $$
For $x>0$, simple guessing shows that $3.1<x<3.2$. But how can we account for $x<0$?
Through guess and check
$x=\dfrac{22}{7}$
We know apriori, $x\approx 3.16 \Longrightarrow \lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor = 28$, so setting $k=28$ yields $x=22/7$
You can verify this yourself:
$$
\tfrac{22}{7}\lfloor\tfrac{22}{7}\lfloor \tfrac{22}{7}\lfloor \tfrac{22}{7}\rfloor\rfloor\rfloor =88
$$
Hint
Let $x=\frac p q\in[k,k+1)$ irreductible and positive so $88\times \frac q p\in\mathbb N$ hence $$p\in\{1,2,4,8,11,22,44,88\}$$ and we have
$$k^4\leq x \lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor = 88\leq (k+1)^4$$ so $k=3$ and then what's the possible values of $x$? Repeat the same reasoning for $x$ negative.
If $0\le x<3$, then the expression is $<3^4<88$. So in the positive case, $k=\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor\ge 27$ and also $k=\frac{88}x\le \frac{88}3$, i.e. $k\in\{27,28,29\}$. Trying $x=\frac{88}{28}=\frac{22}{7}$ we verify that we have found a solution, whereas $\frac{88}{27}$ and $\frac{88}{29}$ don't work. (Since $x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor$ is strictly decreasing (in the domain $x\ge1$) we might not even fully test other values once we have found or otherwise guessed a solution).
To deal with negative $x$, we might substitue $x$ with $-y$ and investigate $$y\lceil y\lceil y \lceil y\rceil\rceil\rceil=88$$ with $y>0$ instead. Once again, for $0<y\le 3$, the left hand side is $\le 81<88$, hence $y>3$, but then the left hand side is $\ge 64 y>192>88$.
There isn't a reason to neglect negative values, but note that if $x=-3-\epsilon$ with $\epsilon$ small ($0\lt \epsilon \lt \frac 1{37}$) you have successively $$\lfloor x \rfloor=-4$$ $$\lfloor -4(-3-\epsilon) \rfloor=12$$ $$\lfloor 12(-3-\epsilon) \rfloor=-37$$$$-37(-3-\epsilon)\gt111$$ And if $x=-3$ the value is $81$, so you can't attain the value $88$ with a negative value of $x$.
If there is a rational solution for $x$, we know that its numerator must divide $88$ evenly. This leaves us with $1$, $2$, $4$, $8$, $11$, $22$, $44$, and $88$ as possible choices of numerator for $x$.
It turns out that $x=\frac{22}{7}$. You can check all other divisors of $88$ to verify that none of them work as a numerator besides $22$.
$$\mathrm{88}={x}\lfloor{x}\lfloor{x}\lfloor{x}\rfloor\rfloor\rfloor \\ $$ $$\lfloor{x}\rfloor=\lfloor\sqrt[{\mathrm{4}}]{\mathrm{88}}\rfloor=\left\{-\mathrm{4},\mathrm{3}\right\} \\ $$ $${if};\:\lfloor{x}\rfloor=-\mathrm{4} \\ $$ $$\mathrm{88}={x}\lfloor{x}\lfloor-\mathrm{4}{x}\rfloor\rfloor \\ $$ $$\lfloor-\mathrm{4}{x}\rfloor=\lfloor\sqrt[{\mathrm{3}}]{\left(-\mathrm{4}\right)^{\mathrm{2}} \left(\mathrm{88}\right)}\rfloor\notin{Z}^{-} \\ $$ $${if};\:\lfloor{x}\rfloor=\mathrm{3} \\ $$ $$\mathrm{88}={x}\lfloor{x}\lfloor\mathrm{3}{x}\rfloor\rfloor \\ $$ $$\lfloor\mathrm{3}{x}\rfloor=\lfloor\sqrt[{\mathrm{3}}]{\left(\mathrm{3}\right)^{\mathrm{2}} \left(\mathrm{88}\right)}\rfloor=\mathrm{9} \\ $$ $$\mathrm{88}={x}\lfloor\mathrm{9}{x}\rfloor \\ $$ $$\lfloor\mathrm{9}{x}\rfloor=\lfloor\sqrt{\left(\mathrm{9}\right)\left(\mathrm{88}\right)}\rfloor=\mathrm{28} \\ $$ $${x}=\frac{\mathrm{88}}{\mathrm{28}}=\frac{\mathrm{22}}{\mathrm{7}} \\ $$
