How can I solve: ${\left [{x+1}\over2\right]}={x-1\over 3}$?

Question

How should the following equation be solved? $${\left [{x+1}\over2\right]}={x-1\over 3}$$ where $[a]$ is the integer part of the number.

Answer 1

First, we know that $\displaystyle \frac{x-1}{3}$ is an integer. So we can write $x = 3k + 1$ for some integer $k$.

Now substitute:

$$k+1+\left[\frac{k}{2}\right] = \left[\frac{3k+2}{2}\right] = k$$

$$\left[\frac{k}{2}\right] = -1$$

So $k = -1$ or $k = -2$. Thus, the solutions for $x$ are $\{ -2,-5 \}$.

Answer 2

The use of the floor function may be confusing because "traditional" means don't work.

But for one thing, we know that the left hand side is an integer, hence so is the right hand side, i.e. $x$ is of the form $3k+1$ with $k\in\mathbb Z$. And we have the estimates $y-1<[y]\le y$, hence $$ \frac{3k+2}{2}-1<k\le \frac{3k+2}{2}.$$ From the left inequality, $k<0$, from the right inequality $k\ge -2$. So there are only few cases left to test: $x=-5$ and $x=-2$.