IMO 2023 | PROBLEM 1 - SOLUTION - 8 July 2023

Problem 1.
Determine all composite integers $n>1$ that satisfy the following property: if $d_1, d_2, \ldots, d_k$ are all the positive divisors of $n$ with 1=d₁<d₂< ... <d_k=n , then d_i divides d_i+1+d_i+2 for every $1 \leqslant i \leqslant k-2$ .

proposed by Santiago Rodriguez, Colombia

Solution
Observe that we have
$d_i | d_{i+1}+d_{i+2} , 1 \leq i \leq k-2$ ,
then put $i=k-2$ then
$d_{k-2} | d_{k-1}+d_{k}=d_{k-1}+n$
but
$d_{k-2} |n \Longrightarrow d_{k-2} | d_{k-1} .$
Again putting $i=k-3$ we get
$d_{k-3} | d_{k-2}+d_{k-1}$
but on other hand
$d_2=\frac{n}{d_{k-1}} | \frac{n}{d_{k-2}} =d_3 , d_2|d_3+d_4 \Longrightarrow d_2|d_4.$
So $d_{k-3}|d_{k-1} \Longrightarrow d_{k-3}|d_{k-2} .$
Now using Induction we get
$d_1|d_2|d_3.......d_k$ .
Hence $n=p^{k-1}$ is the solution.