Here's the details of the problem:
Problem 1. Determine all composite integers $n>1$ that satisfy the following properties: if $d_1, d_2,\cdot \cdot \cdot ,d_k$ 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 $1\leq i \leq k-2$
My solution for this problem starts with the fact that $$d_{k-2}\: \vert \:d_{k-1} + d_{k}$$ but we also have $d_k=n$, therefore $$d_{k-2} \:\vert\: d_{k-1}$$ and let $$xd_{k-2}=d_{k-1}$$
Note that if $x$ is composite, then $x$ has divisor $a$ such that $1 <a <x$, then $ad_{k-2}\:\vert\:n$. But this is a contradiction, because $$d_{k-2}<ad_{k-2}<d_{k-1}$$ Therefore, $x=p$ where $p$ is a prime number, and we know $$d_{k-1}\:\vert\: n$$ Let $n=yd_{k-1}$. Similarly from the above conclusion, $y$ cannot be composite and $y$ is a prime. More precisely, $x=y=p$ (because if $y$ is a different prime $q$ then $qd_{k-2}$ also becomes a devisor of $n$)
We also know that $$d_{k-3}\:\vert\: d_{k-2}+d_{k-1}=d_{k-2}+pd_{k-2}$$ and since $d_{k-3}\:\vert\:n=p^{2}d_{k-2}$, $d_{k-3} \:\vert\: pd_{k-2}$ which means $$d_{k-3} \:\vert\: d_{k-2}$$ and as above, $$pd_{k-3}=d_{k-2}$$
Continuing as above we notice that all divisors should be of the form$$(1,p,p^2,\cdot \cdot \cdot, p^k)$$
Therefore all numbers which satisfy the above condition are numbers $n$ such that $n=p^k$ where $p$ is a prime number
I want to ask: Is my solution correct? Because all the solutions given to the above problem are done in a way different to mine.
Well, I wrote the same solution at the IMO and got 2 points. My team leader told me that the claim $d_k=pd_{k-1}$ is wrong. Don't know why because $n=p^k$ will definitely have its divisors $p$ times the previous one.