This is what I get too. Also, it looks like all non-prime peculiars can be represented as [imath]n=p(p^2-p-1)[/imath] where [imath]p[/imath] is prime. At least all the ones under 99,999 look this way (but I have no clue how to prove this) :
[math]\begin{array}{rcrcr}
15 &=& 3 &\times& 5 \\
95 &=& 5 &\times& 19\\
287 &=& 7 &\times& 41\\
1199 &=&11 &\times& 109\\
4607 &=& 17 &\times& 271\\
23519 &=& 29 &\times& 811\\
28799 &=& 31 &\times& 929\\
\end{array}[/math]
We call a positive integer [imath] n[/imath] peculiar if, for any positive divisor [imath] d[/imath] of [imath] n[/imath] , the integer [imath] d(d + 1)[/imath] divides [imath] n(n + 1).[/imath]
Lemma: If [imath] n[/imath] is peculiar and has at most two prime divisors then [imath] n\in \{p\, , \,p(p^2-p-1)\}[/imath] and [imath] p^2-p-1[/imath] is prime.
Proof: If [imath] p\,|\,n[/imath] then [imath] n=p\cdot m.[/imath] In case [imath] m=1[/imath] we have [imath] p(p+1)=n(n+1)\,|\,n(n+1)[/imath] and [imath] 1\cdot (1+1)= 2\,|\,n(n+1)[/imath] so primes are peculiar. Let us therefore assume that [imath] n[/imath] is not prime and [imath] m>1.[/imath]
Let [imath] m=q[/imath] be prime and [imath] n=p\cdot q[/imath] a product of two primes. In case [imath] p=q[/imath] we get
[math]\begin{array}{lll}
p(p+1)\,|\,n(n+1)&=p^2(p^2+1)=p^4+p^2=p^3(p+1)-p^3+p^2\\
&=p^3(p+1)-p^2(p+1)+2p^2\\
&=p^3(p+1)-p^2(p+1)+2p(p+1)-2p\\
&=p^3(p+1)-p^2(p+1)+2p(p+1)-2(p+1)+2\\
&\Rightarrow 0\equiv 2\bmod{p+1}
\end{array}[/math]which is impossible since [imath] p>1.[/imath] Therefore [imath] p\neq q[/imath] and
[math]\begin{array}{lll}
q(q+1)\,|\,n(n+1)=pq(pq+1)&\Longrightarrow (q+1)\,|\,p(pq+1)=p^2(q+1)-p^2+p\\
&\Longrightarrow (q+1)\,|\,p(p-1)
\end{array}[/math]and for symmetry reasons also [imath] (p+1)\,|\,q(q-1).[/imath] Thus [imath] r\cdot(q+1)=p(p-1)[/imath] and [imath] s\cdot(p+1)=q(q-1)[/imath] for some positive integers [imath] r,s\in \mathbb{Z}.[/imath] If [imath] p=2[/imath] then [imath] (q+1)\,|\,2\cdot(2-1)=2[/imath] which is impossible for [imath] q>1.[/imath] We may thus assume that we have odd primes [imath] 3\leq p <q[/imath] and [imath] q\,\nmid\,(p+1).[/imath] Hence [imath] q\,|\,s[/imath] or [imath] q\cdot t=s[/imath] and [imath] t\cdot(p+1)=q-1[/imath] for some positive integer [imath] t\in \mathbb{Z}.[/imath]
If [imath] p\,\nmid\,(q+1)[/imath] then [imath] p\,|\,r,[/imath] say [imath] p \cdot u=r.[/imath] Thus
[math]0<u=\dfrac{p-1}{q+1}=\dfrac{p-1}{tp+t+2}\in \mathbb{Z}[/math]which implies first [imath] t=0[/imath] and then [imath] s=0[/imath] which is a contradiction. We therefore have [imath] p\,|\,(q+1),[/imath] say [imath] p\cdot k=q+1,[/imath] i.e. [imath] r(q+1)=rpk=p(p-1)[/imath] for some positive integer [imath] k\in \mathbb{Z}[/imath] and
[math]p=rk+1 \text{ and }q=t(rk+2)+1=rk^2+k-1 .[/math]If [imath] t\geq k[/imath] then [imath] rk^2+k-1=t(rk+2)+1\geq rk^2+ 2k+1[/imath] which isn't possible for positive [imath] k.[/imath] Hence we may assume [imath] t<k[/imath] or [imath] -k<-t.[/imath] Now
[math]\begin{array}{lll}
0&=rkt+2t-rk^2-k+2=r(kt-k^2)-(k-2t-2)\\
r&=\dfrac{k-2t-2}{k(t-k)}=\dfrac{2t+2-k}{k(k-t)}>0\\
0&< 2t+2-k\text{ and thus }t<k<2(t+1)\\
k&\in \{t+1,\ldots,2t+1\}\, , \,k-t\in \{1,\ldots,t+1\}\\
r&=\dfrac{2t+2-k}{k(k-t)}\geq 1\\
0&<k(k-t)\leq 2t+2-k < t+2< k+2\\
0&<k-t< 1+\dfrac{2}{k}
\end{array}[/math]Since both, [imath] t[/imath] and [imath] k[/imath] are positive, and [imath] 0<t<k,[/imath] the smallest possible number is [imath] k=2.[/imath] On the other hand, if [imath] k\geq 2,[/imath] then [imath] k-t \in ]0,2[.[/imath] and so [imath] k=t+1. [/imath] We have then in all cases
[math]r=\dfrac{2t+2-k}{k(k-t)}=\dfrac{2t+2-t-1}{t+1}=1[/math]and thus
[math]q=p(p-1)-1=p^2-p-1 \text{ prime. }[/math]as demanded.
So if I didn't make any mistakes, this reduced the problem to ...
If [imath] n [/imath] is peculiar then it has at most [imath] 2 [/imath] prime factors.
The original problem should be easier once we know what peculiar numbers look like.
