User talk:ArtismScrub/Number classifications

According to your definition, no number on your list, and indeed no natural number whatsoever, is empowered - for example, 72 is divisible by 12, but not by 122. If you only meant the condition for primes, this is the same as which aren't powers.

Your condition for when \(a^bc^d\) is empowered is wrong as well - take \(a=b=c=3,d=2\). \(a,c\) are non-powers, neither of \(b,d\) divides the other, yet \(a^bc^d=3^5\) is not empowered.

There are no Steinhaus primes other than 3, since \(n^n-1\) is always divisible by \(n-1\). Hence there are no S-M prime pairs other than 3, 5.

In general, if \(b\) is odd, then \(a^b+1\) is divisible by \(a+1\). It follows that if \(n^n+1\) is a prime, \(n\) can't have any odd factors other than 1, so it's a power of 2, \(n=2^k\). Then again \(k\) can't have an odd factor other than 1, so \(k=2^l\) and \(n=2^{2^l},n^n+1=2^{2^{l+2^l}}+1\), so every Moser prime is a Fermat prime. Hence it is an open problem whether there are infinitely many of those. LittlePeng9 (talk) 08:29, February 16, 2018 (UTC)