OFFSET
1,2
COMMENTS
Conjecture: For k > 1, k^(m + 1) == -1 (mod m) has an infinite number of positive solutions.
Conjecture: For k > 1, if f(1) = p(1) equals one of the prime factors of k^2 + 1, p(i+1) equals one of the prime factors of k^(f(i)+1) + 1 greater than p(i), f(i+1) = f(i)*p(i+1), then k^(f(i) + 1) == -1 (mod f(i)) for all integers i. (Especially in this sequence, k = 4, so {f(i)} can be 17, 4097, 4298117633, ...) - Jinyuan Wang, Nov 16 2019
PROG
(Magma) [n + 1: n in [0..5000000] | Modexp(4, n + 2, n + 1) eq n ];
(PARI) isok(m) = Mod(4, m)^(m+1) == -1; \\ Jinyuan Wang, Nov 16 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
Juri-Stepan Gerasimov, Nov 06 2019
EXTENSIONS
a(13)-a(25) from Giovanni Resta, Nov 08 2019
STATUS
approved