On odd perfect numbers $q^k n^2$ satisfying $n^2 - q^k = 2^r t$

Question

Let $N = q^k n^2$ be an odd perfect number with special prime $q$, satisfying $$n^2 - q^k = 2^r t$$ where $r \geq 2$ and $\gcd(2,t)=1$.

We could prove that:

(1) $2^r t > 2n$. (We can modestly improve this to $$2^r t > \frac{3373n^2}{3375},$$ since per this answer to a related MSE question, we have the lower bound $\sigma(n^2)/q^k \geq {3^3} \times {5^3} = 3375$, which implies that $$q^k < \frac{2n^2}{3375}.)$$

(2) $n^2 - q^k$ is not a square.

Our question is:

Must it necessarily be the case that $n > \max(2^r, t)$?

That is, can we rule out the following cases?

(A) $t > n > 2^r$

(B) $2^r > n > t$

I think that, without further constraints, you can not rule out the cases you state, as all the conditions stated can be satisfied in each of those cases — Juan Moreno, Jun 04 '21 at 13:40
Would you mind sharing more details about your thoughts on this problem @JuanMoreno, by posting an answer? — Jose Arnaldo Bebita Dris, Jun 05 '21 at 04:03

score 2 · Answer 1 · answered Jun 05 '21 at 12:45

2

It is asked if necessarily $n>\max(2^r,t)$, providing certain constraints. This constraints can be satisfied if $n<\max(2^r,t)$, so further constraints are needed.

For instance, consider $n=675$, $q=13$, $k=5$, $r=2$, $t=21083$. All the constraints are satisfied (excepting $N$ being some odd perfect number, which if satisfied would be a breakthrough discovery), but $t>n>2^r$.

answered Jun 05 '21 at 12:45

Juan Moreno

229

Thank you for your answer. I am actually interested in proving $n < q^k$ (and therefore, ruling out $q^k < n$) generally from the constraint $$n^2 - q^k = 2^r t$$ where $r \geq 2$ and $\gcd(2, t)=1$, @JuanMoreno. – Jose Arnaldo Bebita Dris Jun 06 '21 at 06:53
Your supposed example for $t > n > 2^r$ actually contradicts the following requirement for the factor $n$, which is implied by the following estimates: (1) $q^k < n^2$ (Dris, 2012); and (2) $N = q^k n^2 > {10}^{1500}$ (Ochem and Rao, 2012). These two estimates imply $n > {10}^{375}$. – Jose Arnaldo Bebita Dris Jun 06 '21 at 06:59
Anyway, I will be interested in seeing a (specific) example where $q^k < n$, even if it is not true that $n > {10}^{375}$. (Of course!) – Jose Arnaldo Bebita Dris Jun 06 '21 at 07:01
That is, I seek (specific) examples for the conditions $t > n > 2^r$ or $2^r > n > t$, which satisfies the constraint $q^k < n$. – Jose Arnaldo Bebita Dris Jun 06 '21 at 07:12
I got the following example from this answer to a closely related MSE question:
$$n = 9, q = 5, k = 1, r = 2, t = 19$$

$$t > n > 2^r$$

It does seem to be the situation that $t > n > 2^r$ is the problematic case.
– Jose Arnaldo Bebita Dris Jun 06 '21 at 07:31
I have unaccepted your answer, @JuanMoreno, in view of some recent findings pointing to the possibility that $t=1$. (Please refer to this answer1 and answer2.) – Jose Arnaldo Bebita Dris Oct 01 '21 at 10:32
It seems that your proof of $t=1$ is flawed, as pointed out by @mathlove – Juan Moreno Oct 01 '21 at 11:24
Let me work it out for the next couple of weeks, @JuanMoreno. – Jose Arnaldo Bebita Dris Oct 01 '21 at 12:53

On odd perfect numbers $q^k n^2$ satisfying $n^2 - q^k = 2^r t$

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