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

For prime number $n$ , $k = 2$ and exists no $i$ .

Consider $n$ is not a prime number, i.e. $k \geq 3$ .

Since $d_{k - 2}$ is a divisor of $d_{k}$ ,

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

Similarly,

d_{k - 3} | d_{k - 2} + d_{k - 1} \Rightarrow d_{k - 3} | d_{k - 2}

Inductively,

1 = d_{1} | d_{2} | \cdot \cdot \cdot | d_{k - 2} | d_{k - 1} | d_{k} = n

Consider $n$ has more than one prime factor, i.e. $n = p_{1}^{s_{1}} p_{2}^{s_{2}} \cdot \cdot \cdot p_{t}^{s_{t}}$ , where $p_{i}$ is prime number, and $p_{1} < p_{2} < \cdot \cdot \cdot < p_{t}$ and $t \geq 2$ .

Then $d_{2} = p_{1}$ and there exists $i \geq 3$ such that $d_{i} = p_{2}$ .

d_{2} | d_{3} | . . . | d_{i} \Rightarrow p_{1} | p_{2}

Contradictory to the assumption that $p_{2}$ is prime.

Therefore, $n$ with more than one prime factor is not eligible.