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Determine all positive integers $N$ to which the following applies (i) $N$ has more than three different positive integer divisors and (ii) for all divisors $p,q$ of $N$ with $1 < p < q < N$, also $q-p$ is a divisor of $N$. 

$N$ will have 2 as a divisor. Because if $N$ only had odd divisors the difference q-p would be even and it cannot divide $N$ as $N$ only has odd divisors. Contradiction - 2 must be a divisor.

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closed as off-topic by Namaste, Peter Taylor, lulu, GoodDeeds, heropup Sep 6 '18 at 23:17

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    $\begingroup$ Any thoughts? Can you, say, exhibit a few numbers that work? Maybe spot a simple pattern? $\endgroup$ – lulu Sep 6 '18 at 12:53
  • $\begingroup$ Ruling out 1 and N itself I ran through some factor combinations and got e.g.30, 36, 48, 120 that worked out. What makes even more difficult for me is that the factors themselves can be composite. $\endgroup$ – Parzifal Sep 6 '18 at 13:24
  • $\begingroup$ $30$ works? But $2,15$ divide $30$ and $13=15-2$ does not. $\endgroup$ – lulu Sep 6 '18 at 13:31
  • $\begingroup$ Similarly, $2,18$ divide $36$ but $16$ does not. I'll let you confirm that your other numbers are wrong as well. $\endgroup$ – lulu Sep 6 '18 at 13:31
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    $\begingroup$ The difference q-p will be even and it cannot divide N as N only has odd divisors ... So 2 must be a divisor. $\endgroup$ – Parzifal Sep 6 '18 at 14:19
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Let $n\in\mathbb{N}$ such that

(i) $n$ has at least four different divisors (i.e at least two different divisors apart from $1$ and $n$)

(ii) for all divisors $p,q$ of $n$ with $1<p<q<n$, $q-p$ is also a divisor of $n$.

Claim 1: $2$ divides $n$.

Assume to the contrary that $2$ does not divide $n$. Let $p,q$ be divisors of $n$ such that $1<p<q<n$ (they exist by (i)). Then $p$ and $q$ are odd, so $q-p$ is even and a divisor of $n$ by (ii), so $n$ is even which is a contradiction. Thus $2$ divides $n$.

Claim 2: If $n$ is not a power of $2$, then $3$ divides $n$.

Assume to the contrary that $3$ does not divide $n$. Let $p$ be a prime divisor of $n$. Then $p\geq5$. By (ii) and induction, $p-2k$ is a divisor of $n$ for all $k\in\{1,\ldots,(p-1)/2\}$ ($p$ and $2$ are divisors, so $p-2$ is, so $p-4$ is, so $\ldots$). So $p-2((p-1)/2-1)=3$ is a divisor of $n$ which is a contradictions. This $3$ divides $n$.

Claim 3: $n$ cannot have any prime divisors apart from $2$ and $3$.

Assume to the contrary that $n$ has a prime divisor that is not $2$ or $3$. Let $p_1,\ldots,p_\ell$ be all distinct prime divisors of $n$ that are not $2$ or $3$. Then $p:=2p_1\cdots p_\ell$ is a divisor of $n$ and so is $2p-3$ by (ii). Neither $2$, nor $3$, nor any of the $p_i$ divides $2p-3$, so $2p-3$ has a prime divisor that is not in the set $\{2,3,p_1,\ldots,p_\ell\}$ which is a contradiction. Thus $n$ cannot have any prime divisors apart from $2$ and $3$.

So now you know that $n$ is of the form $2^u3^v$. I'll leave it to you to show that $6,8,12$ are the only options.

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  • $\begingroup$ Thank you very much for the comprehensive and well written explanation. Yet I will need some time to work through your solution. $\endgroup$ – Parzifal Sep 6 '18 at 16:07

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