# For $s, t \in \{-\lfloor (n-1)/2 \rfloor, \dots, -1, 0, 1, \dots, \lfloor n/2 \rfloor \}$, $n \in \mathbb{Z}_{ > 0}$, show $s-t \neq \ell n$

This is an intermediate step to a problem that I don't know how to prove.

Added 4/29/2019: (Abstract) algebra is not my strongest subject, but if that is necessary to tackle this problem, I'll take that solution. Solutions that don't require a lot of (abstract) algebra machinery would be preferred.

For the record, I am not certain that this is true.

For any $$s, t \in \{-\lfloor (n-1)/2 \rfloor, \dots, -1, 0, 1, \dots, \lfloor n/2 \rfloor \} = F_n$$ with $$n \in \mathbb{Z}_{ > 0}$$ fixed, show that $$s-t \neq \ell n$$ for any $$\ell \in \mathbb{Z}\setminus\{0\}$$.

My attempt: Suppose $$n$$ is odd. Then $$F_n = \left\{\dfrac{-(n-1)}{2}, \dots, -1, 0, 1, \dots, \dfrac{n-1}{2} \right\}$$ I attempt to proceed by induction.

If $$n = 1$$, we have $$F_1 = \{0\}$$, and clearly $$s - t = 0$$ cannot equal a non-zero multiple of a positive integer.

Suppose $$n = k$$ and the claim holds for $$s, t \in F_k$$.

Then, what I've observed is that $$F_{k+1} = F_k \cup \left\{\dfrac{-(k+1-1)}{2}, \dfrac{k+1-1}{2} \right\} = F_k \cup \left\{\dfrac{-k}{2}, \dfrac{k}{2} \right\}$$ and at this point, I am stuck.

• Since $\max F_n - \min F_n=\lfloor n/2\rfloor+\lfloor (n-1)/2\rfloor=n-1,$ $F_n$ consists of $n$ consecutive integers, so it contains no distinct $t,s$ with $s-t$ divisible by $n$. – Alex Ravsky May 1 at 14:59

There is no need using induction. Note that if $$n$$ is odd, then $$\lfloor\frac{n}{2}\rfloor=\frac{n}{2}-\frac{1}{2}$$ and $$\lfloor \frac{n-1}{2}\rfloor=\frac{n-1}{2}$$ and if $$n$$ is even $$\frac{n}{2}\rfloor=\frac{n}{2}$$ and $$\lfloor \frac{n-1}{2}\rfloor=\frac{n-1}{2}-\frac{1}{2}$$. Hence for all $$n$$ and for all $$s,t\in F_{n}$$ we find that $$|s-t|\leq|\lfloor \frac{n}{2}\rfloor+\lfloor \frac{n-1}{2}\rfloor|=|\frac{n}{2}+\frac{n-1}{2}-\frac{1}{2}|=|n-1|.$$ Hence there is no $$l\in\mathbb{Z}\setminus\{0\}$$ such that $$s-t=ln$$.