One evident pattern is that after 10^1 + 1 = 11 and 10^2 + 1 = 101, there are no further primes (at least up through 10^200 + 1). Can it be true that for every positive integer b >= 2, there is a integer N(b) > 0 such that n >= N(b) => b^n + 1 is not prime ??? Or is this not plausible? If it *is* plausible, is it subject to vast generalization (plausibly)? —Dan Marc LeBrun wrote: -----
On Sep 25, 2017, at 5:30 PM, James Davis <lorentztrans@gmail.com> wrote:
Thanks! That saves me a week.
Then perhaps I can waste you some more time? So square-free, the exceptions pop out! By eyeball it seems like 11^2 is a divisor when n = an odd multiple of 11. And 7^2 divides n=21 and 63, and there's even a 13^2 for n=39 (no doubt all these are related to n=3 being 7 11 13?) What is the smallest n divisible by a cube? By two distinct squares? By any square that's relatively prime to 1001? -----