Any natural number multiple of (12 +i), (22 + i), (34 + i) and (56+ i) is a valid base for pseudoprimality of 561 in Z(i). 40 is the order of these bases (mod 561 ).

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## Comments

## Carmichael numbers 561 (contd)

Let ((12+i)^40-1)/561 = a +ib .Then (-12 - i), (1 - 12i) and (-1+ 12i) all yield a + ib when subjected to the same operation as above. a - ib is yielded by the conjugates of the above four Gaussian integers when subjected to the above operation.

## Carmichael numbers - pari code

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.

## Application of failure functions - indirect primality test

This application is valid for not only x^2+1 but also for other quadratic polynomials including non-monic ones. Members can refer to earlier message pertaining to the case x^2 + 1 for the algorithm.

## Application of failure functions - indirect primality test

This application is valid for not only x^2+1 but also for other quadratic polynomials including non-monic ones. Members can refer to earlier message pertaining to the case x^2 + 1 for the algorithm.

## Application of failure functions - indirect primality test

This application is valid for not only x^2+1 but also for other quadratic polynomials including non-monic ones. Members can refer to earlier message pertaining to the case x^2 + 1 for the algorithm.

## pseudoprimality of powers

We can use pari to find the bases for which powers are pseudoprimes. Example: 25 is pseudoprime to the bases 7, 18 and 24.

## Isomorphism - an example

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

## Isomorphism - an example

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

## Isomorphism?

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.

## Isomorphism - an example

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

## Which isomorphism?

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!

## Which isomorphism?

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!

## which isomorphism?

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!

## Message system

Message system does not work =o(

## messages...

…seem to work for me (I see your message, do you see this?)

## message system

Thanks Joe,

Now the messages are visible again!

The search not…

## A property of exponential functions.

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.

## A property of exponential functions.

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.

## A property of exponential functions.

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.