I'm currently working on odd perfect numbers, to be precise on a proof (german language) of Rudolf Steuerwald. I have two questions regarding this.
Euler proved that any odd perfect number $n$ has to be of the form $$n=p^\alpha \prod_{i=1}^n q_i^{b_i}$$ where
- $p,q_1,...,q_n$ are distinct odd primes,
- $p\equiv\alpha\equiv1 \ (\text{mod} \ 4)$,
- $b_i\equiv0 \ (\text{mod} \ 2)$ for all $i\in \{1,...,n\}$.
Steuerwald proved that not all the $b_i$ can be $2$, what means that $\exists i\in \{1,...,n\}: b_i\geq4$.
First question (p. 69/70)
Steuerwald used a Lemma: All prime factors of $\frac{p+1}{2}$ have to be a $q_i$. Proof: We can use \begin{equation} \frac{p^{\alpha+1}-1}{p-1}=2\cdot\frac{p^{\frac{\alpha+1}{2}}-1}{p-1}\cdot\frac{p^{\frac{\alpha+1}{2}}+1}{p+1}\cdot\frac{p+1}{2} \end{equation} where all factors are integers, because $b_i\equiv0 \ (\text{mod} \ 2)$ for all $i\in \{1,...,n\}$ In addition, due to the fact that $n$ is perfect, we have \begin{equation} \frac{p^{\alpha+1}-1}{p-1}\prod_{i=1}^n\frac {q_i^{b_i+1}-1}{q_i-1}=2p^\alpha \prod_{i=1}^n q_i^{b_i} \end{equation} where the first factor on the left is $\equiv2 \ (\text{mod} \ 4)$ while all others are odd. This proves the Lemma.
I understand both equations but not their connection. Can you help me?
Second question (p. 71):
It's said: (As is well known) a number $q_i^2+q_i+1$ has only prime factors, which are $\equiv1 \ (\text{mod} \ 6)$ or $3$. For me this is not obvious and I wasn't able to proof it. Can you help me?
Thank you very much in advance. Sorry for my bad English!
\begin{equation} \frac{p^{\alpha+1}-1}{p-1}\prod_{i=1}^n\frac {q_i^{b_i + 1}-1}{q_i-1}=2p^\alpha \prod_{i=1}^n q_i^{b_i}. \end{equation}
– Jose Arnaldo Bebita Dris May 05 '18 at 11:25