Van der Pol's identity for the sum of divisors and a quartic polynomial equation for odd perfect numbers

In Touchard (1953) it is mentioned that the sum of divisors $$\sigma(n)$$ satisfies the following recurrence relation ($$n>1$$):

$$n^2(n-1) = \frac{6}{\sigma(n)} \sum_{k=1}^{n-1}(3n^2-10k^2)\sigma(k)\sigma(n-k)$$

Substituting in this equation an odd perfect number $$n = \sigma(n)/2$$ we find that it is a root of the quartic polynomial:

\begin{align} f(x) &= x^4-x^3+12a_0x^2-60a_1x+60a_2 \\ &= (x^3+(n-1)x^2+(12a_0+n^2-n)x+12a_0n+n^3-n^2-60a_1)(x-n) \end{align}

where

$$a_i = \sum_{k=1}^{\frac{n-1}{2}}k^i \sigma(k)\sigma(n-k),\text{ } i=0,1,2$$

We have:

$$n^3(n-1) = 3 \sum_{k=1}^{n-1} (3n^2-10k^2)\sigma(k) \sigma(n-k)$$ and by symmetry of $$\sigma(k)\sigma(n-k)$$ in $$k,n-k$$ we get:

$$= 3 \sum_{k=1}^{(n-1)/2} (6n^2-10k^2-10(n-k)^2)\sigma(k) \sigma(n-k)$$ which might be simplified to:

$$=-12n^2a_0 -60a_2+60na_1$$ which proves that the odd perfect number $$n$$ is a root of the polynomial $$f(x)$$ above.

However, computations with Sagemath (it takes a few seconds to do the computation) seem to suggest, that for an odd number $$m$$, we have

$$f(m) > 0, \text{for } m>1 \text{ odd}$$

This observation would maybe prove, that there are no odd perfect numbers. I am aware that $$\sigma(k)\sigma(n-k)$$ is kind of like a black box to handle, but it would be nice, if maybe for some special cases where $$m$$ is odd the last inequality could be proven.

It might be noticed that it seems like for $$n>1$$, odd the polynomial $$f(x)$$ is irreducible over the rational numbers.

Here are the first polynomials:

n, f(n,t), f(n,n)>0?
3 t^4 - t^3 + 36*t^2 - 180*t + 180 f(n,n)>0 ? True
5 t^4 - t^3 + 228*t^2 - 1860*t + 3300 f(n,n)>0 ? True
7 t^4 - t^3 + 696*t^2 - 7920*t + 20160 f(n,n)>0 ? True
9 t^4 - t^3 + 1548*t^2 - 22500*t + 72900 f(n,n)>0 ? True
11 t^4 - t^3 + 2940*t^2 - 51600*t + 204600 f(n,n)>0 ? True
13 t^4 - t^3 + 4956*t^2 - 102360*t + 478920 f(n,n)>0 ? True
15 t^4 - t^3 + 7752*t^2 - 185760*t + 1004400 f(n,n)>0 ? True
17 t^4 - t^3 + 11376*t^2 - 307620*t + 1900260 f(n,n)>0 ? True
19 t^4 - t^3 + 16020*t^2 - 481980*t + 3309420 f(n,n)>0 ? True
21 t^4 - t^3 + 22080*t^2 - 735120*t + 5559120 f(n,n)>0 ? True

Is there a irreducibility criterion which might be applied to all ($$n=3,\ldots,21$$) of these example polynomials?

Edit: The odd perfect number is also a root of the polynomial:

$$g(x) = x^4-x^3-9A_0x^2+30A_2$$

where

$$A_i = \sum_{k=1}^{n-1}k^i \sigma(k)\sigma(n-k),i=0,2$$

But numerical computations suggest, that:

$$\gcd(f(x),g(x))=1$$

Can this last claim about the $$\gcd$$ be proven?

An answer to the question: It would be nice if someone can spot the error in this answer, otherwise it would follow that there are no odd perfect numbers...:

Let $$A_i = \sum_{k=1}^{n-1} k^i \sigma(k)\sigma(n-k),\text{ } i=0,1,2,3$$.

Using the symmetry of $$\sigma(k)\sigma(n-k)$$ in $$n,k$$ for $$A_i$$ we get:

$$A_0 = 2a_0$$ $$A_1 = na_0$$ $$A_2 = a_0n^2-2a_1n+2a_2$$ $$A_3 = n^3a_0-3n^2a_1+3na_2$$

On the other hand we can evaluate $$A_i$$ with Eisenstein series (https://mathoverflow.net/questions/372766/is-there-a-similar-formula-like-ramanunjans-eisenstein-series-identity-for-su):

$$A_0 = \tfrac5{12}\sigma_3(n)-\tfrac12n\sigma(n)+\tfrac{\sigma(n)}{12} = \tfrac5{12}\sigma_3(n)-\tfrac12n \cdot 2n +\tfrac{2n}{12}$$

$$A_1 = \frac{5n}{24}\sigma_3(n) - \frac{6n^2-n}{24}\sigma(n) = \frac{5n}{24}\sigma_3(n) - \frac{6n^2-n}{24}2n$$

$$A_2 = \frac{n^2}{8}\sigma_3(n) - \frac{4n^3-n^2}{24}\sigma(n) = \frac{n^2}{8}\sigma_3(n) - \frac{4n^3-n^2}{24}2n$$

$$A_3 = \frac{n^3}{12}\sigma_3(n) - \frac{3n^4-n^3}{24}\sigma(n) = \frac{n^3}{12}\sigma_3(n) - \frac{3n^4-n^3}{24}2n$$

We can solve the set of equations in $$a_i,A_i$$ for $$a_i$$ and get:

$$a_0 = 1/2 A_0$$ $$a_2=\frac{A_{2}-nA_1}{2} +na_1$$

We can further substitute for $$A_i$$ the last equations in $$\sigma_3(n)$$ and then substitute these $$a_i$$ in $$f(x)$$.

We get (here are the calculations done with Sagemath, I hope there is no bug in this): $$0 = f(n) = n^4 + 6A_0n^2 - n^3 + 30A_1n + 30A_2$$ $$= -5/2(12n^2 - 5\sigma_3(n) - 2n)n^2$$

Hence:

$$0 = 12 \, n^{2} - 2 \, n - 5 \, \sigma_{3}(n)$$

But $$\sigma_3(n)>n^3+1$$ and we get the contradiction for $$n>1$$:

$$5(n^3+1) < 5 \sigma_3(n)= 12n^2-2n < 5(n^3+1)$$

• "Substituting in this equation an odd perfect number" : This is confusing - no odd perfect number is known. Commented Sep 25, 2020 at 7:16
• What is "it" in "it is a root of the quartic polynomial"? I thought it was $n$, but you have $n$ there separately, in addition to $x$. And your $a_i$ depend on $n$ too, so this is not really a quartic equation for $n$. Commented Sep 25, 2020 at 7:30
• The coefficients are quite complicated structured. I do not think that there is a hope to prove irreducibility and therefore the desired contradiction Commented Sep 25, 2020 at 7:37
• If $(x-n)\mid f(x)$ how can $f$ be irreducible? Commented Sep 27, 2020 at 9:01
• @TheSimpliFire If I understood it right, the OP tries a contradiction proof : If we can show that $f(x)$ is irreducible and on the other hand inserting a perfect odd number $n$ gives $f(n)=0$, this would be a contradiction showing that there is no perfect odd number. Commented Sep 27, 2020 at 9:30