Disclaimer: this problem came from USAMTS: https://www.usamts.org/Tests/Problems_31_3.pdf
The contest has ended., in case there is any doubt.
Problem: Find all squarefree integers whose divisors $d_1 < d_2 < · · · < d_k$ satisfy $d_i − d_{i−1}|n$ for all $2 \leq i \leq k$.
My thinking is if $d_i − d_{i−1}|n$ then $d_i − d_{i−1}$ is in $d_1, d_2,… d_k$. Then $d_2-d_1 < d_2$ is a divisor so $d_2-d_1 = d_1, d_2 = 2d_1$. Obviously $d_1 = 1$ and $d_2 = 2$ since $1 | n$.
Then $(d_3 – d_2 = d_3 – 2$. if $d_3 – 2 = d_2 $ then $d_3 = 4$. If $d_3 – 2 = d_1 $ then $d_3 = 3$
Powers of 2 seem to work, but not sure how to prove if other cases don't work.
It seems we can make some decent progress but I am stuck here..
Best Answer
We must have $d_1=1.$ So $d_2-1$ must be divisor less than $d_2$ so $d_2-1=1,$ or $d_2=2.$
In general, if $p$ is a prime divisor of $n$ then let $d$ be the previous divisor. Then $p\not\mid d$ and $p-d$ must be a divisor of $n$ and likewise $\frac{n}{d}-\frac{n}{p}=\frac{(p-d)n}{pd}=(p-d)\frac{n}{pd}$ must be a divisor of $n.$
But we can show that $\gcd(p-d,pd)=1.$ So if $d\neq p-1$ then $(p-d)\frac{n}{pd}$ is not square free, so it cannot be a divisor of $n.$
So the only way for $p$ to be added is if $p-1$ is also a divisor.
In particular, we must start $1,2,3,\dots,$ and the next cannot be $5$ since $5\neq 3+1.$ So the next must be six, and the next value must be a prime, and the only prime can be $6+1=7.$
So we start $1,2,3,6,7,\dots,14,\dots, 21,\dots,42,\dots$ If there were any primes added to this in the $\dots,$ then the smallest such $p$ must have $p-1$ in $7,14,21,42.$ But the only option is $p=43,$ because $7+1,14+1,21+1$ are all non-prime.
So the sequence must start:
$$1,2,3,6,7,14, 21,42,\dots$$
We can stop at $n=42.$ Or we can continue with $p=43.$ But then we have that there is not prime $p$ so that $p-1=43d_1$ where $d_1\mid 42.$ So there can be no larger $n.$
So the only values are $n=1,2,6=2(2+1),42=6(6+1),1806=42(42+1).$