Conjecture about natural number satisfying $ m(n)^k+1\space\mid\space n^{2k}+1 $ Let $m(n)$ be the greatest proper divisor of $n$. Is there any number $n≥2$ not of the form $p$ or $p^3$ for $p$ prime that satisfies
$$
m(n)^k+1\space\mid\space n^{2k}+1
$$
for all natural numbers $k$?
I haven't found any of them, but I reduced it to the case where $n=pq$ for $p$, $q$ prime and with $p<q<p^2$. In this case, we must also have:
$$
q^k+1\space\mid\space p^{2k}+1
$$
But I didn't manage to break it further down; somehow the condition appears to be extremely hard to manage, although it seems pretty strong.
 A: If $n$ is not prime then $n=pm$ where $m=m(n)$ and $p$ is the smallest prime dividing $n$. Then the condition implies
$$ m+1 \mid m^2p^2+1 $$
But
$$ m+1 \mid (m+1)(m-1)p^2 = m^2p^2-p^2 $$
hence
$$ m+1 \mid m^2p^2+1-(m^2p^2-p^2) = p^2+1 $$
and we must have $p^2\ge m$.
If $n\ne p^3$ then $m<p^2$. So $m$ cannot have two prime factors $\ge p$, but nor can it have any prime factors $<p$ since $p$ is the smallest prime factor of $n$. Hence $m$ must be prime.
Let $r>n$ be a prime with $\left(\frac{m}{r}\right)=-1$.$^\dagger$
Then by Euler's criterion
$$
m^{\frac{r-1}{2}} \equiv -1 \pmod r \\
r \mid m^{\frac{r-1}{2}}+1
$$
But
$$
n^{r-1}+1 \equiv 2 \pmod r
$$
and hence
$$
m^{\frac{r-1}{2}}+1 \not\mid n^{r-1}+1
$$
Hence if $n$ is not of the form $p$ or $p^3$ with $p$ prime, then the condition cannot be satisfied for all $k$.
$\dagger$ Given $m$ prime we can always find a prime $r>n$ with $\left(\frac{m}{r}\right)=-1$. Let $b$ be any quadratic nonresidue mod $m$. By the Chinese Remainder Theorem we can find $r_0$ with $r_0 \equiv 1 \pmod {4}$ and $r_0 \equiv b \pmod m$. Then by Dirichlet's theorem there is a prime $r>n$ with $r\equiv r_0 \pmod {4m}$, and by quadratic reciprocity
$$
\left(\frac{m}{r}\right) = \left(\frac{r}{m}\right) = \left(\frac{b}{m}\right) = -1
$$
