Odd perfect numbers, questions on a proof of Steuerwald 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!
 A: For the second question, we first have $(3q+1)^2+3q+1+1=9q^2+6q+1+3q+1+1=3(3q^2+3q+1)$, showing that there is always at least one factor that is not $3$ (for $q\ge 1$).
For the other factors, note the approach listed at https://math.stackexchange.com/a/2187430/86846 specifically that we should take our $q^2+q+1$ and substitute using $$p-1=q\\\to q^2+q+1=(p-1)^2+p-1+1\\=p^2-p+1.$$  Then we have the same numeric form as the one linked (using $N=p$), and the conclusion is the same, that any prime $u$ where $u\mid p^2-p+1=q^2+q+1$ must have $6\mid u-1$.
More specifically, since $(p+1)(p^2-p+1)=p^3+1$ means that $p^3\equiv -1\mod u\to p^6\equiv 1\mod u$ for any prime $u\mid p^2-p+1$, we then have that the order of $p$ modulo $u$ must divide $6$, but if this order is anything less than $6$ we get a contradiction.  Since the group of units modulo $u$ is of order $u-1$, we get the desired result, that $6\mid u-1\to u=6k+1$.
The form $q^2+q+1$ must be changed since using $(q-1)(q^2+q+1)=q^3-1$ does have a factor of the form $6n-1$ (i.e., $6n-1\mid q-1$) when $q^2+q+1\equiv 1\mod 6$.
