# On a conjectured relationship between the least prime factor and the Euler prime of an odd perfect number

(Note: This question has been cross-posted to MO.)

Let $\sigma$ be the classical sum-of-divisors function. For example, $$\sigma(6)=1+2+3+6=12={2}\cdot{6}.$$

If $\sigma(N)=2N$ and $N$ is odd, then $N$ is called an odd perfect number. Currently, the problem of the existence of odd perfect numbers is still open.

Euler proved that an odd perfect number $N$ must have the form $$N=q^k n^2,$$ where $q$ is prime with $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

Dris  showed that the inequality $$\sigma(q^k) \leq \dfrac{2n^2}{3}$$ holds, and subsequent improvements were given by other authors (Dris, Luca (2016), Chen, Chen (2012), Broughan, Delbourgo, Zhou (2013), and Chen, Chen (2014)).

Let $p_1$ be the least prime factor of the odd perfect number $N$ with Euler prime $q$. It is relatively easy to show that $p_1 \neq q$. Suppose to the contrary that the Euler prime $q$ is also the least prime factor of the odd perfect number $N$. Since $\sigma(q) \mid \sigma(q^k) \mid \sigma(N) = 2N$, we get that $$\dfrac{q+1}{2} = \dfrac{\sigma(q)}{2} \mid N,$$ so that $(q+1)/2$ divides $N$. If $(q+1)/2$ is prime, then this will contradict the assumption that $q$ is the least prime factor of $N$ (as $(q+1)/2 < q$). If $(q+1)/2$ is composite, then $(q+1)/2$ will have a prime factor (necessarily smaller) that likewise divides $N$. Again, this is a contradiction. QED

Therefore, $p_1 < q$.

Here is my question:

Will it be possible to either prove or rule out the following equality, using the current state of knowledge that we have on odd perfect numbers? $$\dfrac{q+1}{2} = {p_1}$$

• The calculation that I can get, but doesn't solve nothing about your question is: expand using Newton's Binomial theorem to get from $(q+1)^k= 2^kp_1^k$ then $$q^k+\sum_{j=0}^{k-1}\binom{k}{j}q^j=2^kp_1^k,$$ and from this (since one wants to use the inequality $\sigma(q^k)\leq \frac{2n^2}{3}$) then multiply by $q$, substract $1$ and dividing by $q-1$ one gets $$\sigma(q^k)+\frac{q}{q-1}\sum_{j=0}^{k-1}\binom{k}{j}q^j=\frac{q(2^kp_1^k)-1}{q-1},$$ and now one can use the inequality $\sigma(q^k)\leq \frac{2n^2}{3}$, and try deduce something (!). – user243301 Sep 28 '16 at 16:20
• @user243301, that is a good start. I invite you to write that out as an actual answer. – Jose Arnaldo Bebita-Dris Sep 28 '16 at 16:28
• I understand that is only a contribution as a comment, is not an answer. I was writting those simple calculations because perhaps you can deduce something useful about the size or from the first of those identities something when you divide, or combining with different factors. Very thanks much. – user243301 Sep 28 '16 at 17:05