<s>[INST] One particular equation involving floor function I've got this equation to solve
$ \left\lfloor{\frac {x + 1}{x+7}}\right\rfloor  = \left\lfloor{\frac {x }{x+3}}\right\rfloor $
It's not very hard to solve it if using properties of rational functions but I wonder if it can be solved using only arithmetics. Anyone got idea?
 [/INST] If $x \ge 0$ then $0 < \frac {x+1}{x+7} < 1$ and $0 \le \frac x{x +3} < 1$ so $[\frac{x+1}{x+1}] = [\frac x{x+3}] = 0$.  So all $x \ge 0$ are solutions.
If $x < 0$ is a matter of solving $[\frac {y - 1}{y-7}] = [\frac y{y- 3}]$ for $y = - x > 0$.
If $y = 1$ then $[\frac {y - 1}{y-7}] = 0 \ne [\frac y{y- 3}] = [- 1/2] = -1$.
Otherwise $\frac {y-1}{y-7} = 1 + \frac 6{y-7}$ and $\frac y{y-3} = 1 + \frac 3{y-3} $ 
so $[\frac {y - 1}{y-7}]=[\frac y{y- 3}]\iff [\frac 6{y-7}] =[\frac 3{y-3}] $
$[\frac 6{y-7}] = k \iff k \le \frac 6{y-7}< k + 1 \iff \frac 6{k}+7\ge y >\frac6{k+1} +7 $
Likewise $[\frac 3{y-3}] = k \iff \frac 3k + 3 \ge y > \frac 3{k+1} +3$
So $y >0 $ $[\frac 3{y-3}]= [\frac 6{y-7}] \iff \min(\frac 6k + 7,\frac 3k + 3) \ge y > \max (\frac 6{k+1} + 7, \frac 3{k+1} + 3)$.
If $k = 0$ then this is true if $y > 13$ and so $x < -13 \implies [\frac {x+1}{x+7} ]=[\frac {x}{x+3}] = 1$
If $k \ge 1$ then this is true if $\frac 3k + 3 \ge y > \frac 6{k+1} + 7 \ge \frac 6{2k} + 7 = \frac 3k +7$ which is impossible.
If $k = -1$ then to be true, $0 \ge y $ which is impossible.
If $k = -2$ then $\frac 32 \ge y > 1$ allows $[\frac{x+1}{x+7}] =[\frac{x}{x+3}] = -1$.
if $k = -3$ then $\frac 3{k} + 3 \ge y  > \frac 6{k + 1} + 7 > \frac 6{2k} + 7 = \frac 3k + 7$ which is impossible.
So solution is $x < -13; -\frac 32 \le x < -1$; $x \ge 0$
</s>