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=d_1<d_2<\cdots<d_k=n

, then

d_i

divides

d_{i+1}+d_{i+2}

for every

1\le i\le k-2

Step 2 of 3: A second prime factor creates a gap

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In plain words

If n had two different prime factors, the smallest divisors would run through powers of the smaller prime until the larger prime appears, and pairing $d_i\leftrightarrow n/d_i$ reproduces the same gap among the largest divisors.

d_1=1,\ldots,d_{e+1}=p^e,\ d_{e+2}=q

Detailed analysis

Suppose $n$ has at least two distinct prime factors, and let $p<q$ be the two smallest ones. Let $p^e$ be the largest power of $p$ dividing $n$ with $p^e<q$ (so $e\ge1$ ). No divisor of $n$ can lie strictly between $p^e$ and $q$ , so the smallest divisors of $n$ are $1,p,\ldots,p^e,q$ . Since $d_id_{k+1-i}=n$ for all $i$ , the divisors just below $n$ mirror this pattern: $d_k=n,\ d_{k-1}=n/p,\ldots,d_{k-e}=n/p^e,\ d_{k-e-1}=n/q$ .