# If $N = q^k n^2$ is an odd perfect number with $k > 1$ and $N = \frac{q(q+1)}{2} \cdot d$ for some $d > 1$, then what is $d$?

By this answer, we know that every odd perfect number $N = q^k n^2$ can be written in the form $$N = \dfrac{q(q+1)}{2} \cdot d$$ where $d > 1$. (That is, an odd perfect number $N = q^k n^2$ is a nontrivial multiple of the triangular number $$T(q) = \dfrac{q(q+1)}{2},$$ where $q$ is the Euler prime of $N$.)

If $k=1$, then it is easy to show that $$d = D(n^2)$$ where $D(n^2) = 2n^2 - \sigma(n^2)$ is the deficiency of the non-Euler part $n^2$.

Here is my question:

If $N = q^k n^2$ is an odd perfect number with $k > 1$ and $N = \frac{q(q+1)}{2} \cdot d'$ for some $d' > 1$, then what is $d'$?

• if you just setup the quadratic in q, you will get the condition for d' by requiring the discriminant D to be a square. if I am not mistaken, D=1+8N/d'. And this severely limits the possible values for d' since for all triangular numbers, 1+8N is a square. Jan 24, 2017 at 20:50
• @user25406, thank you for your comment. Fleshing it out now into an actual answer with more details included. I hope you do not mind! =) Jan 24, 2017 at 20:56
• @user25406, I think your claimed result is trivial. Please check out the answer that I have posted below. Jan 24, 2017 at 21:08
• I think you misunderstood my comment. I meant that the only solutions are the trivial ones d'=N to get 1+8=3^2. I did say "severely limits the possible values" for d'. Jan 24, 2017 at 21:20
• Hold on. Are you claiming that for all triangular numbers $N$, $1+8N$ is a perfect square? Jan 24, 2017 at 21:24

This is more to the spirit of my expected answer for the original question, rather than the one presented in my other answer.

So we have an odd perfect number $N = q^k n^2 = {q(q+1)/2}\cdot{d'}$ (with $k>1$), where $d' > 1$ by Slowak's 1999 result.

Thus, $\frac{2N}{q(q+1)} = \frac{\sigma(N)}{q(q+1)} = d'$, so that $$d' = \dfrac{\sigma(n^2)\sigma(q^k)}{q(q+1)} = {q^{k-1}}\cdot{\dfrac{\sigma(n^2)}{q^k}}\cdot{\dfrac{\sigma(q^k)}{q+1}}.$$

Now, I have shown elsewhere that $$\dfrac{\sigma(n^2)}{q^k}=\frac{D(n^2)}{\sigma(q^{k-1})},$$ so that we obtain $$d' = {q^{k-1}}\cdot{\dfrac{D(n^2)}{\sigma(q^{k-1})}}\cdot{\dfrac{\sigma(q^k)}{q+1}}.$$ Finally, we have $$d' = \bigg(\dfrac{\sigma(q^k)}{\sigma(q)}\bigg)\cdot\bigg(\dfrac{D(n^2)}{I(q^{k-1})}\bigg)$$ or equivalently, $$d' = {q^{k-1}}\cdot\bigg(\dfrac{I(q^k)}{I(q)I(q^{k-1})}\bigg)\cdot{D(n^2)},$$ where $I(x)=\sigma(x)/x$ is the abundancy index of the (positive) integer $x$.

Note that these last two formulas reduce to the case $d = D(n^2)$ when $k=1$, as expected.

As hinted by user25406 in a comment, setting up the quadratic in $q$, we obtain $${d'}q^2 + {d'}q - 2N = 0.$$

In order for this equation to have (positive) integer solutions for $q$, we need the discriminant $$D_1 := {d'}^2 + 8{d'}N$$ to be a perfect square.

This gives that $$D_1 = {{d'}^2}\cdot(1 + 4q(q+1)) = \left({d'}(2q+1)\right)^2$$ is a perfect square, which is trivial.

To clarify another comment, rewrite the equation $${d'}q^2 + {d'}q - 2N = 0$$ as $$q^2 + q - \dfrac{2N}{d'} = 0.$$
The discriminant is given by $$D_2 := 1 + \dfrac{8N}{d'}.$$
user25406's (unproved) claim is that this is a perfect square only when $d' = N$.
• Yes, for $d'=N$, the discriminant $D_2$ is a perfect square. But are you sure that $d'=N$ is the only value of $d'$ that makes $D_2$ a perfect square? Note that, per Ochem and Rao's result on a lower bound for the magnitude of an odd perfect number, $N > {10}^{1500}$. Jan 25, 2017 at 0:44
• @user25406, my point in the initial version of my answer was that: Given the discriminant $D_2 = 1 + (8N/d')$, if you substitute $N = {q(q+1)/2}\cdot{d'}$, you get $D_2 = 1 + 4q(q+1) = (2q+1)^2$, which is indeed a perfect square. Note that this value for the discriminant no longer depends on $N$ or $d'$, but only on the value of $q$. Jan 25, 2017 at 0:48
• Any way, the reason why I am highly skeptical about your (claimed) proof for $d' = N$ is that it will imply $q = 1$, which would contradict the fact that $q$ is the Euler prime (that is, $q$ is a prime number and ought to satisfy $q \equiv 1 \pmod 4$). I invite you to write out your complete proof for $d' = N$, as it will prove the Descartes-Frenicle-Sorli conjecture on odd perfect numbers (that is, that $k=1$ ought to hold). Jan 25, 2017 at 0:52