TITLE: Solve the system $ x \lfloor y \rfloor = 7 $ and $ y \lfloor x \rfloor = 8 $.
QUESTION [14 upvotes]: Solve the following system for $ x,y \in \mathbb{R} $: \begin{align} x \lfloor y \rfloor & = 7, \\ y \lfloor x \rfloor & = 8. \end{align}

It could be reducing to one variable, but it is not that simple.
Another idea is substitution, but it didnt lead me anywhere.
This was taken from a competition, and it’s tricky.
From what I noticed, systems with two or more variables involving "greatest integer" function are rarely seen, probably too difficult to analyze...

This was obtained with the help of Wolfram Alpha: (I swapped places of 7 and 8 by mistake, so it is not exactly the same system; anyway, the answer should not rely on software tools, of course)

REPLY [3 votes]: Let $x=m+\xi$, $y=n+\eta$ with $m$, $n\in{\mathbb Z}$, and $\xi$, $\eta\in[0,1[\ $. Then we have to solve $$(m+\xi)\>n=7,\qquad(n+\eta)\>m=8\ .\tag{1}$$
(i) If $m\geq1$, $n\geq1$ then $(1)$ implies
$$m=\left\lfloor{7\over n}\right\rfloor,\qquad n=\left\lfloor{8\over m}\right\rfloor\ .\tag{2}$$ The following table contains in the first row possible values of $n$, in the second row the resulting values of $m$ according to $(2_1)$, and in the third row the resulting values of $n$ according to $(2_2)$:
$$\matrix{
n&&1&2&3&4&5&6&7&\geq8 \cr
m:={\displaystyle\left\lfloor{7\over n}\right\rfloor} &&7&3&2&1&1&1&1&0\cr
{\displaystyle\left\lfloor{8\over m}\right\rfloor} &&1&2&4&8&8&8&8&-\cr}$$
A value $n$ in the first row is usable if the corresponding entry in the third row coincides with $n$. This leads to the usable pairs $(m,n)\in\bigl\{(7,1), (3,2)\bigr\}$, from which one computes the candidates
$$(x,y)\in\left\{\left(7,{8\over7}\right), \left({7\over2},{8\over3}\right)\right\}\ .$$
(ii) If $m':=-m\geq1$, $n':=-n\geq1$ then $(1)$ implies
$$(m'-\xi)\>n'=7,\qquad (n'-\eta)\>m'=8\ ,$$
or
$$n'=\left\lceil{8\over m'}\right\rceil,\qquad m'=\left\lceil{7\over n'}\right\rceil\ .\tag{3}$$
The following table contains in the first row possible values of $m'$, in the second row the resulting values of $n'$ according to $(3_1)$, and in the third row the resulting values of $m'$ according to $(3_2)$:
$$\matrix{ 
m' &&1&2&3&4&5&6&7&\geq8 \cr
n':={\displaystyle\left\lceil{8\over m'}\right\rceil} &&8&4&3&2&2&2&2&1\cr
{\displaystyle\left\lceil{7\over n'}\right\rceil} &&1&2&3&4&4&4&4&7\cr}$$
A value $m'$ in the first row is usable if the corresponding entry in the third row coincides with $m'$. This leads to the usable pairs $(m',n')\in\bigl\{(1,8), (2,4),(3,3),(4,2)\bigr\}$, from which one computes the candidates
$$(x,y)\in\left\{\left(-{7\over8},-8\right), \left(-{7\over4},-4\right),\left(-{7\over3},-{8\over3}\right),\left(-{7\over2},-2\right)\right\}\ .$$
It remains to check that all six candidate pairs $(x,y)$ that we have found indeed satisfy the original equations.