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`|` -> `\mid`, repeated word, and name of MO question
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LSpice
LSpice
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Let $\sigma(n)$ be the sum of the divisors of $n$. Is it always true that if $n$ is odd, that $$n|\sum_{k=1}^{\frac{n-1}{2}}k^2\sigma(k)\sigma(n-k)?$$$$n\mid\sum_{k=1}^{\frac{n-1}{2}}k^2\sigma(k)\sigma(n-k)?$$

I have checked this up to $n=100$, and I suspect there's some simple argument using for why this should be true, probably using Eisenstein series identities but I'm not seeing it.

The motivation from this is from thisthe old Mathoverflow question Mathoverflow questionVan der Pol's identity for the sum of divisors and a quartic polynomial equation for odd perfect numbers, where this term appears as the constant in the polynomials generated which for an integer $n$, $n$ is an odd perfect number if and only if $n$ is a root of the $n$th polynomial there. It seems unlikely that the method used there will work as any sort of viable attack on that problem, since they have a very tough set of quartic polynomials they would need to prove are irreducible, and one has very little control over the behavior of the coefficients. But if the above relationship is true, then this at least can be replaced with a similar looking cubic polynomial which still seems unlikely to be helpful, but at least marginally more plausible as an angle of productive attack since then if there is an integer root, the other two roots have to be quadratics. And the coefficient of the $x^2$ will always be $-1$, so the sum of the roots is pretty restricted.

Let $\sigma(n)$ be the sum of the divisors of $n$. Is it always true that if $n$ is odd, that $$n|\sum_{k=1}^{\frac{n-1}{2}}k^2\sigma(k)\sigma(n-k)?$$

I have checked this up to $n=100$, and I suspect there's some simple argument using for why this should be true, probably using Eisenstein series identities but I'm not seeing it.

The motivation from this is from this old Mathoverflow question, where this term appears as the constant in the polynomials generated which for an integer $n$, $n$ is an odd perfect number if and only if $n$ is a root of the $n$th polynomial there. It seems unlikely that the method used there will work as any sort of viable attack on that problem, since they have a very tough set of quartic polynomials they would need to prove are irreducible, and one has very little control over the behavior of the coefficients. But if the above relationship is true, then this at least can be replaced with a similar looking cubic polynomial which still seems unlikely to be helpful, but at least marginally more plausible as an angle of productive attack since then if there is an integer root, the other two roots have to be quadratics. And the coefficient of the $x^2$ will always be $-1$, so the sum of the roots is pretty restricted.

Let $\sigma(n)$ be the sum of the divisors of $n$. Is it always true that if $n$ is odd, that $$n\mid\sum_{k=1}^{\frac{n-1}{2}}k^2\sigma(k)\sigma(n-k)?$$

I have checked this up to $n=100$, and I suspect there's some simple argument for why this should be true, probably using Eisenstein series identities but I'm not seeing it.

The motivation from this is from the old Mathoverflow question Van der Pol's identity for the sum of divisors and a quartic polynomial equation for odd perfect numbers, where this term appears as the constant in the polynomials generated which for an integer $n$, $n$ is an odd perfect number if and only if $n$ is a root of the $n$th polynomial there. It seems unlikely that the method used there will work as any sort of viable attack on that problem, since they have a very tough set of quartic polynomials they would need to prove are irreducible, and one has very little control over the behavior of the coefficients. But if the above relationship is true, then this at least can be replaced with a similar looking cubic polynomial which still seems unlikely to be helpful, but at least marginally more plausible as an angle of productive attack since then if there is an integer root, the other two roots have to be quadratics. And the coefficient of the $x^2$ will always be $-1$, so the sum of the roots is pretty restricted.

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JoshuaZ
JoshuaZ
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Divisibility relation with a specific sum of divisors

Let $\sigma(n)$ be the sum of the divisors of $n$. Is it always true that if $n$ is odd, that $$n|\sum_{k=1}^{\frac{n-1}{2}}k^2\sigma(k)\sigma(n-k)?$$

I have checked this up to $n=100$, and I suspect there's some simple argument using for why this should be true, probably using Eisenstein series identities but I'm not seeing it.

The motivation from this is from this old Mathoverflow question, where this term appears as the constant in the polynomials generated which for an integer $n$, $n$ is an odd perfect number if and only if $n$ is a root of the $n$th polynomial there. It seems unlikely that the method used there will work as any sort of viable attack on that problem, since they have a very tough set of quartic polynomials they would need to prove are irreducible, and one has very little control over the behavior of the coefficients. But if the above relationship is true, then this at least can be replaced with a similar looking cubic polynomial which still seems unlikely to be helpful, but at least marginally more plausible as an angle of productive attack since then if there is an integer root, the other two roots have to be quadratics. And the coefficient of the $x^2$ will always be $-1$, so the sum of the roots is pretty restricted.