Hello cndman Originally Posted by

**cndman** Hey Guys, I have two problems that have to do with the factor theorem. I've been stumped on them for sometime now and can no longer put them off as my assignment is due soon.

1) Use the factor theorem to show that if 2^p - 1, where p does not equal 3, is a prime number, then p is neither divisible by 4 or divisible by 3. {Alternatively, prove that if p is divisible by 4 or 3, the 2^p - 1 is divisible by some number other than +/- itself or +/-1.}

Known, factor theorem: p(x) = q(x)(x-a) +r(x)

I've tried setting 2^p - 1 = 0, but I get P log 2 = 0, I guess if someone could give me a hint or push me in the right direction I'd greatly appreciate it. thanks

We use two results here:$\displaystyle x^2-1=(x-1)(x+1)$ and $\displaystyle x^3-1=(x-1)(x^2+x+1)$

First, suppose that $\displaystyle p =4m, m \in \mathbb{N}, m\ge1$. Then$\displaystyle 2^p-1 = 2^{4m}-1$$\displaystyle =(2^{2m})^2-1$

$\displaystyle =(2^{2m}-1)(2^{2m}+1)$

$\displaystyle \Rightarrow 2^p-1$ has a factor $\displaystyle 2^{2m}-1\ge 3$ and a factor $\displaystyle 2^{2m}+1\ge 5$. Now suppose that $\displaystyle p = 3m, m \in \mathbb{N}, m\ge2$. Then$\displaystyle 2^p-1 = 2^{3m}-1$$\displaystyle =(2^{m})^3-1$

$\displaystyle =(2^{m}-1)((2^{m})^2+2^m+1)$

$\displaystyle \Rightarrow 2^p-1$ has a factor $\displaystyle 2^{m}-1\ge 3$ and a factor $\displaystyle (2^{m})^2+2^m+1\ge 21$

Grandad