Factors of $2^p + 1$, p prime Has the following conjecture been proved or a counterexample found?
If $p$ is a prime not equal to 3, then $2^p + 1$ is a square-free integer.
 A: I think this has been neither proved nor disproved. I've heard at least one mathematician opine that it is probably false (though I don't know of a supporting heuristic).
A: Choose an odd prime $q>3$ and let the least positive integer $h$ such that $2^{h} \equiv -1$ (mod $q^{2}$), if there is such an integer. Then $2^{h}-1$ is relatively prime to $q,$ so $2$ has multiplicative order $2h$ in $\mathbb{Z}/q^{2}\mathbb{Z}.$ Hence if $2^{n} \equiv -1$ (mod $q^{2},$ we must have $n \equiv h$ (mod 2h) and $h$ divides $n.$ If $n$ is prime, this forces $h =p$ 
as $q >3$ so $h \neq 1.$ Hence to answer the question in the negative, we need to find odd primes $p$ and $q$ so that $2$ has multiplicative order $2p$ in $\mathbb{Z}/q^{2}\mathbb{Z}.$ I can't see any obvious reason why there should not be such primes $p$ and $q,$ though I can imagine they might be rare, for the following reason. Note that we would require $p|q-1$ in this case. If we choose a pair of odd primes $p$ and $q$ with
$p|q-1$ and we choose a random integer $a$ coprime to $q,$ and consider its order in $\mathbb{Z}/q^{2}Z$, we consider the number of elements of order $2p$ in this group
of units. There are $p-1$ elements of order $2p$ in this cyclic group, and a total of $q^{2}-q$ elements in the group. So it is reasonable to consider the probability that $a$ has multiplicative order $2p$ as $\frac{p-1}{q^{2}-q} < \frac{1}{q}.$ In particular, this applies when $a = 2.$
