How do I deal with a floor function is a system of equations?

Question

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?

Answer 1

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 $.

Answer 2

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 $$x = 12p/143, \text{ for $p = 0, 1, 2, \dots , 142$ }; \quad y = 12(x - \lfloor x \rfloor).$$