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%I #21 Sep 08 2022 08:46:24
%S 1,17,4097,7361,85073,658529,3999137,72281281,285143057,628944689,
%T 854112113,1423081169,2561019281,3111576929,4298117633,5921265041,
%U 14224884929,21336998129,34317377233,50723421713,63797137889,144269032049,163834314353,187397322209,212565453281
%N Numbers m that divide 4^(m + 1) + 1.
%C Conjecture: For k > 1, k^(m + 1) == -1 (mod m) has an infinite number of positive solutions.
%C 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
%o (Magma) [n + 1: n in [0..5000000] | Modexp(4, n + 2, n + 1) eq n ];
%o (PARI) isok(m) = Mod(4, m)^(m+1) == -1; \\ _Jinyuan Wang_, Nov 16 2019
%Y Cf. A055685.
%Y Solutions to k^(m + 1) + 1 == -1 (mod m): A296369 (k = 2), A328230 (k = 3).
%K nonn
%O 1,2
%A _Juri-Stepan Gerasimov_, Nov 06 2019
%E a(13)-a(25) from _Giovanni Resta_, Nov 08 2019
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