# 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

1. $p,q_1,...,q_n$ are distinct odd primes,
2. $p\equiv\alpha\equiv1 \ (\text{mod} \ 4)$,
3. $b_i\equiv0 \ (\text{mod} \ 2)$ 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 $$\frac{p^{\alpha+1}-1}{p-1}=2\cdot\frac{p^{\frac{\alpha+1}{2}}-1}{p-1}\cdot\frac{p^{\frac{\alpha+1}{2}}+1}{p+1}\cdot\frac{p+1}{2}$$ 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 $$\frac{p^{\alpha+1}-1}{p-1}\prod_{i=1}^n\frac {q_i^{b_i+1}-1}{q_i-1}=2p^\alpha \prod_{i=1}^n q_i^{b_i}$$ 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!

• For the second question, make a chart of the residues modulo $3$ of $k$ in $k^2+k+1$, for example, if $k\equiv 0\mod 3$ then $k^2+k+1\equiv 1\mod 3$, etc. – abiessu May 4 '18 at 16:56
• The second equation in your first question should be $$\frac{p^{\alpha+1}-1}{p-1}\prod_{i=1}^n\frac {q_i^{b_i + 1}-1}{q_i-1}=2p^\alpha \prod_{i=1}^n q_i^{b_i}.$$ – Jose Arnaldo Bebita-Dris May 5 '18 at 11:25
• To answer your first question: Notice that $\gcd(p, p+1)=\gcd(p,(p+1)/2)=1$, so that all the prime factors of $(p+1)/2$ have to be a $q_i$. – Jose Arnaldo Bebita-Dris May 5 '18 at 11:29
• Additionally, note that, since $p \equiv 1 \pmod 4$, then $(p+1)/2 \equiv 1 \pmod 2$. – Jose Arnaldo Bebita-Dris May 5 '18 at 11:30
• @JoseArnaldoBebitaDris Thank you very much, i got it! – Starter628496 May 5 '18 at 15:19

## 1 Answer

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 $q^2+q+1\equiv 1\mod 6$.