Van der Pol's identity for the sum of divisors and a quartic polynomial equation for odd perfect numbers https://ift.tt/eA8V8J

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?

Thanks for your help!

(Originally asked on MO: https://mathoverflow.net/questions/372476/van-der-pols-identity-for-the-sum-of-divisors-and-a-quartic-polynomial-equation)

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:

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

which would lead to a contradiction.

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

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