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## Latest Messages

Jul 31
Thanks Joe, Now the messages are visible again! The search not...

[P] messages... by jac Jul 28
...seem to work for me (I see your message, do you see this?)

Jul 27
Message system does not work =o(

Jul 27
Ok, $\mathbb{Z}_{31}$ is an additive group of order 31 (and in fact a Galois field since 31 is prime). But I'm interested in the isomorphism you are speaking of. The expression $\frac{2^{5n}-1}{31}$ says me nothing $-$ excuse me!

Jul 27
Ok, $\mathbb{Z}_{31}$ is an additive group of order 31 (and forms in fact a field since 31 is prime). But I am interested which isomorphism you are speaking of. The expression $\frac{2^{5n}-1}{31}$ says me nothing $-$ excuse me!

Jul 27
Ok, $\mathbb{Z}_{31}$ is an additive group of order 31 (and forms in fact a field since 31 is prime). But I am interested which isomorphism you are speaking of. The expression $\frac{2^{5n}-1}{31}$ says me nothing $-$ excuse me!

Jul 27
Let f(n) = a^n + c (a,n and c belong to N, n is not fixed ). Let M_p be a Mersenne prime. If M-p does not exactly divide f(n) for n = 1 to p then M_p does not exactly divide f(n) for any value of n, however large n may be.

Jul 27
Let f(n) = a^n + c (a,n and c belong to N, n is not fixed ). Let M_p be a Mersenne prime. If M-p does not exactly divide f(n) for n = 1 to p then M_p does not exactly divide f(n) for any value of n, however large n may be.

Jul 27
Let f(n) = a^n + c (a,n and c belong to N, n is not fixed ). Let M_p be a Mersenne prime. If M-p does not exactly divide f(n) for n = 1 to p then M_p does not exactly divide f(n) for any value of n, however large n may be.

Jul 27
Since we know that 561 is not a Carmichael number in Z(i) we need to know the code for searching those bases for which 561 is a pseudoprime. Code in pari: {p(n) = ((n + i)^40 - 1)/561}. Incidentally 153 + i is also a valid base for pseudoprimality of 561; needless to say its associates 153 - i etc. are also valid bases.

Jul 27
Pahio, I meant the finite group of remainders (mod 31)-sorry I typed " ïnfinite ".

Jul 25
Hi, BCI1! It's me, Ascold1. Bye!

Jul 25
Deva, can you please explain in detail which is the (infinite) group you mean? I see only the expression $\frac{2^{5n}-1}{31}$. The group $\mathbb{Z}_{31}$ is finite.

Jul 25
(2^(5n) -1)/31 is an infinite group isomorphic with Z_31. Here n belongs to N.