## FERMAT AND MERSENNE NUMBERS CONJECTURE-(2)

14 02 2009

(2)-CONSTRUCTING SOME FUNCTION ZEROS

$f(n)=Mod( \phi(n),\sigma_0(n))$

(2.1) PRIMES:

$\displaystyle f(p)=Mod(p-1,2)=0$, holds for every odd prime $p \in \mathbb{P} -\{2\}$.

$f(2)=1$

(2.2) PRODUCT OF DISTINCT PRIMES NOT 2:

$\displaystyle f(p_1*p_2*...*p_n)=0$, because $\displaystyle \sigma_0(p_i)|\phi(p_i) \rightarrow {2 |(p_i-1)}$, always holds if $\displaystyle p_{i} \neq 2$

With $\displaystyle k$, distinct primes, none of them equal two, it is possible to combine them in $\displaystyle 2^n$, products to find $\displaystyle 2^n$ zeros.

This set of zeros can be described as odd squarefree numbers [1].

(2.3) POWERS OF 2:

$\displaystyle f(2^k)=0\;\rightarrow (k+1)|2^{k-1}$, so $\displaystyle k+1=2^n$ must be a power of 2:

$\displaystyle k=2^n-1=M_n$
A power of 2, is a function zero, iff the exponent is a Mersenne number.

$\displaystyle f(2^{M_n})=f(2^{2^{n}-1})=0$

(2.4) POWERS OF A PRIME:

$\displaystyle f(p^k)=0\;\rightarrow (k+1)|(p-1)*p^{k-1}$, then the zeros can fall into two cases:

(2.4.1) Case: $\displaystyle (k+1)|(p-1)$

Then if $\displaystyle d|(p-1)$, $\displaystyle f(p^{d-1})=0$, and we can built $\displaystyle \sigma_{0}(p-1)$, zeros, one for every divisor of $\displaystyle (p-1)$.

Note that, in the particular case:

$\displaystyle (k+1)=(p-1) \rightarrow{k=p-2}\rightarrow{f(p^{p-2})=0}$.

And $\displaystyle f(p^{p-1})=Mod((p-1)*p^{p-2},p)=0$.

(2.4.2) Case: $\displaystyle (k+1)|p^{k-1}$

This is more general than (2.3):

$\displaystyle f(p^{p^{n}-1})=0$

Archives:

References:

[1]-CRCGreathouse at My Math Forum/Number Theory: Mersenne and Fermat Numbers congruence