IMO 2023 Problem 1

15 Sep, 2024

最近三到五年以来，我一直沉浸在一种长期的，低烈度的倦怠之中。就像一种缓慢而持久的发炎一样，始终没有糟糕到需要解决一下的程度，但就是...很痛苦。

做了很多自救的尝试，最新的一个是打算做点数学竞赛的题目。一方面算是一种brain teasing，让自己变得兴奋起来，或者起码对事情感到一些兴趣，另一方面也是，回到人生比较高光的时刻的状态。

今天的题目是2023年IMO的第一题数论：

Problem 1, 2023 IMO

Determine all composite integers $n > 1$ that satisfy the following property: if $d_{1}, d_{2}, . . ., d_{k}$ are all the positive divisors of n with $1 = d_{1} < d_{2} < \cdot \cdot \cdot < d_{k} = n$ , then $d_{i}$ divides $d_{i + 1} + d_{i + 2}$ for every $1 ⩽ i ⩽ k - 2$ .

Conclusion

The eligible positive integer $n$ must contain exactly one prime factor, i.e. $n = p^{s}$ , $p$ is prime number.

Proof

Neccessity

It's trivial to prove that condition in the problem holds for all $n = p^{s}$ .

Sufficiency

$d_{k - 2} | d_{k - 1} + d_{k} \Rightarrow d_{k - 2} | d_{k - 1}$

Since all factors are ordered in ascending order, $d_{2} = \frac{d_{k}}{d_{k - 1}}, d_{3} = \frac{d_{k}}{d_{k - 2}}, \frac{d_{3}}{d_{2}} = \frac{d_{k - 1}}{d_{k - 2}} \in ℕ \Rightarrow d_{2} | d_{3}$

$d_{2} | d_{3} + d_{4} \Rightarrow d_{2} | d_{4}$

By induction, $\forall i \in {3, 4, \dots, k} : d_{2} | d_{i}$

If $n$ has more than one prime factor (assume ordered ascendingly $p_{1} < p_{2} < \dots$ ), $d_{2} = p_{1}, \exists i \in {3, 4, \dots, k} : d_{i} = p_{2}$ . This is contradictory to the assumption that $d_{2} | d_{i}$

Therefore, $n$ has exactly 1 prime factor.

#math-competition