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 1 of 3: Prime powers: identify the divisors

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In plain words

When n has only one prime factor, its divisors are just the increasing powers of that prime, so the chain condition becomes a single easy divisibility fact.

n=p^r\implies d_i=p^{i-1}

Detailed analysis

If $n=p^r$ for a prime $p$ and $r\ge2$ , its divisors in order are $d_i=p^{i-1}$ for $i=1,\ldots,r+1$ . The required relation $d_i\mid d_{i+1}+d_{i+2}$ becomes $p^{i-1}\mid p^i+p^{i+1}=p^i(1+p)$ , which holds because $i-1\le i$ . Hence every prime power satisfies the property.