How do I deal with a floor function is a system of equations? How would one solve an equation with a floor function in it:
\begin{cases}
  y=12(x-\lfloor x \rfloor) \\ 
  x=12(y-\lfloor y \rfloor) 
\end{cases}
Maybe an algebraic method could be used?
 A: Write $x = m + \alpha, y = n + \beta$ where $m, n$ are integers and $0 \leq \alpha, \beta < 1$. The system is equivalent to the conditions $x = 12 \beta$, $y = 12 \alpha$ and the linear system
$$\begin{align*}
-\alpha + 12\beta &= m, \\
12 \alpha - \beta &= n.
\end{align*}
$$
Solving this system, we find 
$$\alpha = \frac{1}{143}(m + 12n) ,\quad \beta = \frac{1}{143}(12m + n).$$
Therefore the solutions to the original system are all pairs $(x,y)$ with
$$x = \frac{12}{143}(m + 12n) ,\quad y = \frac{12}{143}(12m + n),$$
where $m, n$ are integers satisfying $0 \leq 12m + n, m + 12n < 143$. Moreover it is clear from the original problem that we must have $0 \leq m, n \leq 11$, and that all $m, n$ satisfying these inequalities yield solutions, except $m = n = 11$. Since $m + 12n$ runs through all numbers $0, 1, 2, \dots , 142$, we find in conclusion that the solutions are all pairs $(x,y)$ with
$$x = 12p/143, \text{ for $p = 0, 1, 2, \dots , 142$ }; \quad y = 12(x - \lfloor x \rfloor).$$
A: Let
$x = n+a$,
$y=m+b$
where
$0 \le a, b < 1$.
Then
$m+b
=12a
$,
so
$0 \le m+b < 12$.
Similarly
$n+a
= 12b
$
so
$0 \le n+a < 12$.
This gives a very finite
number of possible cases,
which can be readily checked.
Possibly more could come from
$n+a
= 12 b
= 12 (12a-m)
=144a-12m
$
or
$n+12m
=143a
$.
