Question: 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 \(1 \leqslant i \leqslant k-2.\)
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Solution 1:
Note that the three largest divisors of \(n\) are either $$\left \{ \frac{n}{q}, \frac{n}{p}, n \right \}$$ for distinct prime divisors \(p\) and \(q\) of \(n\), or $$\left \{ \frac{n}{p^2}, \frac{n}{p}, n \right \}$$ for some prime divisor \(p\) of \(n.\)
In the former case we have a contradiction, since
$$\frac{\frac{n}{p} + n}{\frac{n}{q}} = \frac{q(p+1)}{p},$$
obviously not an integer.
So, the three largest divisors of \(n\) are \(\frac{n}{p^2}, \frac{n}{p}\) and \(n\).
With similar reasoning, now consider the fourth largest divisor. It is either \(\frac{n}{q}\) or \(\frac{n}{p^3}\); the former option fails, since
We repeat this reasoning to deduce that \(n\) is just \(p^{k-1}\).
Solution 2:
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.