IMO 2008 N1 #
Let $m, n, p$ be non-negative integers with $m$ odd and $p$ prime. Let $x_1, x_2, …, x_m$ be integers such that $$ x_1^n + px_2 = x_2^n + px_3 = … = x_m^n + px_1. $$ Prove that $x_1 = x_2 = … = x_n$.
Solution #
We follow Solution 2 of the official solution, with some shortcuts made along the way. For $n$ odd, we show that $x_i < x_{i + 2}$ implies $x_{i + 2} < x_{i + 4}$, and similarly $x_i > x_{i + 2}$ implies $x_{i + 2} > x_{i + 4}$. In particular, $p$ only needs to either be odd or equal to $±2$; the above modification makes sure that the solution works even when $p < 0$. Our implementation will also assume $p$ is either odd or $±2$ instead of being prime. For the case $p = ±2$, $n = 2k$, after showing $|x_i^k + x_{i + 1}^k| = 2$, we deduce $x_{i + 1}^k = x_{i - 1}^k$ for some $i$ and then get $x_{i + 2} = x_i$.
Notes #
The original problem considers $m = 3$, but the same method works for any odd $m$. However, the statement is false when $p = 4$ and $m = 3$. A counterexample is given by $(x_1, x_2, x_3) = (7, -5, 1)$ with $n = 2$.
Extra lemmas #
The product of several integers is odd iff each term is odd.
The product of several integers is odd iff each term is odd.
If the product of several natural numbers is one, then each term is one.
If the product of several natural numbers is one, then each term is one.
The natAbs of product is equal to the product of natAbs.
Start of the problem #
A sequence x_1, x_2, …, x_m of integers is called (n, p)-good
if x_i^n + px_{i + 1} = x_j^n + px_{j + 1} holds for any i : ℕ.
Equations
Instances For
Suppose p ≠ 0, and let (x_i) be an (n, p)-good sequence.
If x_{i + j} = x_i, then x_{i + jk} = x_i for any k.
If p ≠ 0 and (x_i) is an (n, p)-good sequence of odd length
such that x_{i + 2} = x_i for some i, then (x_i) is constant.