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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?

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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 $.

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

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  • $\begingroup$ Your answer is more complete than mine. Nice finish. $\endgroup$ – marty cohen Feb 10 '16 at 5:44
  • $\begingroup$ Thanks very much for the very neat and clear solution but would be glad if you can explain how you get " 0≤12m+n,m+12n<143". $\endgroup$ – uranium Feb 10 '16 at 7:06
  • $\begingroup$ That's because the condition on $\alpha$ and $\beta$ is that they belong to $[0,1)$. $\endgroup$ – David Feb 10 '16 at 7:08
  • $\begingroup$ Can you show me how to break the given two equations into 23 set of simple of linear equations from which the 143 points can be found geometrically? thanks $\endgroup$ – uranium Feb 10 '16 at 7:53
  • $\begingroup$ I'm not sure. For a fixed value of $m$, all the solutions lie on one line. The solutions will be located on the $12$ lines corresponding to $m = 0, 1, 2, \dots, 11$. Perhaps you want to take the intersections with the $12$ lines given by $n = 0, 1, 2, \dots, 11$. $\endgroup$ – David Feb 10 '16 at 8:24

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