Find all number $n\in\mathbb{N}$ such that $$n-1 \mid 1 + 10^n + 10^{2n}$$
If $n-1$is prime, it's clear that $n-1\mid 10^{n-2} - 1$ if $n\ne 3, 6$. So, $$0\equiv 1 + 10^n + 10^{2n} \equiv 1 + 10^2 + 10^4\mod n-1\implies n=4, 8, 14, 38$$
But I can't continue after this. I also found that $n = 2, 4, 8, 14, 22, 38, 40, 92, 112, 260, 1652$ works. Is this an open question?
Comment: We rewrite A as:
$A=10^{2n}+10^n+1=(10^n+1)^2-10^n$
Suppose n is even $n=2k$, the we may write:
$A=(10^{2k}+1-10^k)(10^{2k}+1+10^k)$
For $k=1$ we have:
$A=91\times 3\times 37$
For $k=2$ we have:
$A=9901\times 3\times 13\times 37$
We can continue and find infinite factors for $k\in \mathbb N$. Each of these factors can be a candidate for n, i. e we for example can have:
$(n-1)= 91, 9901, 3, 7, 13, 37, . . .$
$n=92, 9902, 4, 8, 14, 38, . . .$
Generally we may write:
$10^{2n}+10^n+1\equiv 10^{2k}+10^k+1=(91, 9901 . . .)\times (3\times 7\times 13\times 37 . . .)$
For example if you take $n=38$ you will get new factors in addition to factors previously found including $n-1=37$ which meets the condition. That is solutions make a recursive relation. Hence the relation can have infinite solutions.