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

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 Commented Jun 4, 2021 at 13:40
• Would you mind sharing more details about your thoughts on this problem @JuanMoreno, by posting an answer? Commented Jun 5, 2021 at 4:03

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 $$t>n>2^r$$.
• 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. Commented Jun 6, 2021 at 6:53
• 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 $n > {10}^{375}$. Commented Jun 6, 2021 at 6:59
• 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!) Commented Jun 6, 2021 at 7:01
• That is, I seek (specific) examples for the conditions $t > n > 2^r$ or $2^r > n > t$, which satisfies the constraint $q^k < n$. Commented Jun 6, 2021 at 7:12
• 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. Commented Jun 6, 2021 at 7:31