That will come later. No, just more a^n=a.
(By the way, I'm thinking I chose a bad offset in talking about this. a^(n+1)=a would be better, because then it transmits to divisors; here I have to write that as "If true for n, it's true for m st m-1|n-1". And subtracting 1 from what I have so far would not be so bad. Oh well. I'll stick with this way for now.)
So I realized 2^n+2 does get us another infinite family of numbers, but, unfortunately, a singularly useless one. The nice thing about having any infinite family of numbers is that the property naturally transmits downard, but not upward. Well, applying this downward transmission to 2^n+2 gets us numbers of the form p^n+1, where p is 3 or 5 mod 8. (You can see that applying this downward transmission to the set of 2^n+2, we get the question, "For which n is -1 a power of 2 (mod n-1)?" And one obvious answer is, "When n-1 is a prime that's 3 or 5 mod 8" (or in fact when it's a power of such a prime). Unfortunately, any divisor of a prime power is itself a power of the same prime, so we can't get anything new from this that way. And we can't even get any more powers of 2 from it because we'd need p to be 7 mod 8 for that (we kind of already have 4, which was how we got this. :P ) In fact primes that are 7 mod 8 you *definitely* can't get this way as 2 is a QR but -1 is not.
The list so far:
2, 3, 4, 5, 2^n+2, p^n+1 for p≡3,5 (8).
-Sniffnoy
(By the way, I'm thinking I chose a bad offset in talking about this. a^(n+1)=a would be better, because then it transmits to divisors; here I have to write that as "If true for n, it's true for m st m-1|n-1". And subtracting 1 from what I have so far would not be so bad. Oh well. I'll stick with this way for now.)
So I realized 2^n+2 does get us another infinite family of numbers, but, unfortunately, a singularly useless one. The nice thing about having any infinite family of numbers is that the property naturally transmits downard, but not upward. Well, applying this downward transmission to 2^n+2 gets us numbers of the form p^n+1, where p is 3 or 5 mod 8. (You can see that applying this downward transmission to the set of 2^n+2, we get the question, "For which n is -1 a power of 2 (mod n-1)?" And one obvious answer is, "When n-1 is a prime that's 3 or 5 mod 8" (or in fact when it's a power of such a prime). Unfortunately, any divisor of a prime power is itself a power of the same prime, so we can't get anything new from this that way. And we can't even get any more powers of 2 from it because we'd need p to be 7 mod 8 for that (we kind of already have 4, which was how we got this. :P ) In fact primes that are 7 mod 8 you *definitely* can't get this way as 2 is a QR but -1 is not.
The list so far:
2, 3, 4, 5, 2^n+2, p^n+1 for p≡3,5 (8).
-Sniffnoy
