The title is spoiler ayan.nmath wrote: Let $n\ge 2$ be an integer. Solve in reals: \[|a_1-a_2|=2|a_2-a_3|=3|a_3-a_4|=\cdots=n|a_n-a_1|.\] Solution. If all are equal, we get an obvious solution. Assume the otherwise. The problem statement is nothing but \[\pm\frac{1}{1}+\pm \frac{1}{2} + \ldots + \pm \frac{1}{n} = 0.\]Pick a prime $p$ between $\frac{n}{2}$ and $n$ (Bertrand's). We will have $p\mid 1\cdot 2\cdot \ldots \cdot (p-1) \cdot (p+1)\cdot \ldots n$. Therefore $n\geq 2p$ which is false. Therefore the only solution is when $a_1 = a_2 = \ldots = k$ for some constant $k$.

Seems like you all had great fun at IMOTC

Yes we did.

Of course we need to show that $\sum_{1 \leq i \leq n} \frac{\varepsilon_i}{i}$ is never zero for $\varepsilon_i \in \{-1, 1 \}$; this is easily done by inducting to prove $\nu_2 \left( \sum_{1 \leq i \leq n} \frac{\varepsilon_i}{i} \right) = \lfloor \log_2 n \rfloor$ using $\nu_2(a+b) \leq \max( \nu_2(a), \nu_2(b) )$ with equality if $\nu_2(a) \neq \nu_2(b)$.

We show that the only solution is $a_1=a_2=\cdots=a_n$, which clearly satisfies. Suppose $|a_1-a_2|=d\ne0$, then we have $d\left(\pm\frac{1}{1}\pm\frac{1}{2}\pm\cdots\pm\frac{1}{n}\right)=0$. By Bertrand's Postulate, there is always a prime $p$ such that $\frac{n}{2}<p\leq n$ if $n\geq2$. Take the inverse of this prime on left side of the equation and the others on the right side. Clearly, the denominator of the RHS is not divisible by $p$, but that of the LHS is. This gives the required contradiction. So, the only solution is $a_1=a_2=\cdots=a_n$, as claimed. $\blacksquare$

The answer is $a_1 = a_2 = \cdots = a_n$. This clearly works. Now we show it is the only solution. It suffices to show that $|a_1 - a_2| = 0$. Suppose otherwise. Let $|a_1 - a_2| = k \ne 0$. We have $|a_i - a_{i + 1}| = \frac ki$ for all $i \in [n]$, where $a_{n+1} = a_1$. Now, for each $i \in [n]$, let $c_i \in \{-1,1\}$ satisfy $a_{i + 1} - a_i = c_i \cdot \frac ki$. We have \[ 0 = a_{n+1} - a_1 = \sum_{i = 1}^{n} (a_{i + 1} - a_i) = \sum_{i=1}^n c_i \cdot \frac ki\] Dividing both sides by $k$ (since $k$ is nonzero) gives that \[ \sum_{i=1}^n \frac{c_i}{i} = 0\] Let $p \in \left( \frac n2 ,n \right]$ be a prime, which exists by Bertrand's Postulate since $n > 1$). Note that for each $i \in [n]$ not equal to $p$, we have $\nu_p \left( \frac{c_i}{i} \right) = 0$ (as there are no other multiples of $p$ in $[1,n]$), but $\nu_p \left( \frac{c_p}{p} \right) < 0$. So the $\nu_p$ of the sum must be less than zero (as we are summing a number with negative $\nu_p$ with numbers of nonnegative $\nu_p$). Hence the sum cannot equal to zero, which is a contradiction. Therefore, the sequence is constant.

Let $a_{n+1}:=a_1$. We have $$(i-1)|a_{i-1}-a_i|=i|a_i-a_{i+1}|\Rightarrow i a_{i+1} \in \left \{a_i+(i-1)a_{i-1}, \ (2i-1)a_i-(i-1)a_{i-1} \right\},$$for all $2\leq i\leq n$. From this it's easy to prove by induction that for every $2<i\leq n+1$ there is a number $c_i$ such that $$a_i=c_i a_2+(1-c_i)a_1 \qquad (1)$$So $a_1=c_{n+1}a_2+(1-c_{n+1})a_1$, i.e. $a_1=a_2$. Now from $(1)$ it follows that $a_i=a_1$ for all $i\leq n$.