Liouville type identity and Elementary Proofs of Modular Identities

Liouville’s Identity:

Let ${f}$ be an even function on integers, we have the identity

$\displaystyle \sum_{\substack{(a, b, x, y) \in \mathbb{N}^{4} \\ ax+by =2n\\ a, b, x, y \text{ odd } }}(f(a+b)-f(a-b))=\sum_{\substack{m \in \mathbb{N} \\ m\mid n \\ n/m \text{ odd }}} m(f(2 m)-f(0))$

This is easy to prove once we know the identity!

Just compare the number of time ${f(k)}$ occurs on both sides. For instance, only even terms ${f(2k)}$ occur.
${f(0)}$ occurs when ${a=b}$ on the LHS and the number of times it occurs is corresponds to divisors ${a | 2n}$ and turns out to be

$\displaystyle \sum_{\substack{m \in \mathbb{N} \\ m\mid n \\ n/m \text{ odd }}} m.$

A generating function proof is to extrapolate the values

$\displaystyle f(0), f(\pm 4), f(\pm 6), \cdots f(\pm 2n).$ to a degree ${2n}$ polynomial, and then compute the LHS in terms of the polynomials.

Choosing ${f(x)=x^2}$ gives

$\displaystyle \sum_{\substack{\ell=1 \\\ell \text { odd }}}^{2 n-1} \sigma(\ell) \sigma(2 n-\ell)=\sum_{\substack{m \in \mathbb{N}\\ m \mid n \\ n / m \text { odd }}} m^{3}.$

We use the following more general identity to prove some modular identities for divisor functions and representation numbers.

Let ${f}$ be any function on integers such that

$\displaystyle f(a, b, x, y)-f(x, y, a, b)=f(-a,-b, x, y)-f(x, y,-a,-b).$

We have

$\displaystyle \sum_{a x+b y=n}(f(a, b, x,-y)-f(a,-b, x, y)+f(a, a-b, x+y, y)$

$\displaystyle -f(a, a+b, y-x, y)+f(b-a, b, x, x+y)-f(a+b, b, x, x-y))$

$\displaystyle = \sum_{d \mid n} \sum_{x<d}(f(0, n / d, x, d)+f(n / d, 0, d, x)+f(n / d, n / d, d-x,-x)$

$\displaystyle -f(x, x-d, n / d, n / d)-f(x, d, 0, n / d)-f(d, x, n / d, 0)).$

The conditions of the theorem are satisfied if for instance we have

$\displaystyle f(a,-b, x, y)=f(-a, b, x, y), \quad f(a, b, x,-y)=f(a, b,-x, y).$

A special case is if ${f(a, b, x, y)=F(a-b, x-y)}$ with ${F(x, y)=F(-x, y)=F(x,-y)}$ .

Then we have the Liouville’s Identity,

$\displaystyle \begin{aligned} &\sum_{a x+b y=n}(F(a-b, x+y)-F(a+b, x-y)) \\ &=\sum_{d \mid n}(d-1)(F(0, d)-F(d, 0))+2 \sum_{d \mid n} \sum_{e<n / d}(F(d, e)-F(e, d)). \end{aligned}$

Example:

a) ${f(a, b, x, y) = xy}$

Check that this satisfies the conditions for the identitty.

Then the LHS equals

$\displaystyle E= \sum_{a x+b y=n} 2xy= 2 \sum_{m=1}^{n-1} \sigma(m) \sigma(n-m).$

The RHS equals

$\displaystyle \sum_{d \mid n} \sum_{x<d}\left(d x+x^{2}-\frac{n^{2}}{d^{2}}\right)=\sum_{d \mid n} \left( d\frac{d(d-1)}{2} + \frac{d(d-1)(2 d-1)}{6} -\frac{n^2}{d^2}\right)$

$\displaystyle =\frac{5}{6} \sigma_{3}(n)+\left(\frac{1}{6}-n\right) \sigma(n)$

Therefore

$\displaystyle \sum_{m=1}^{n-1} \sigma(m) \sigma(n-m)=\frac{1}{12}\left(5 \sigma_{3}(n)+(1-6 n) \sigma(n)\right).$

This also implies

$\displaystyle \sum_{m=1}^{n-1} m \sigma(m) \sigma(n-m)=\frac{n}{24}\left(5 \sigma_{3}(n)+(1-6 n) \sigma(n)\right)$

$\displaystyle f(a, b, x, y)=x y^{3}+x^{3} y$

gives

$\displaystyle \sum_{m=1}^{n-1} \sigma(m) \sigma_{3}(n-m)=\frac{1}{240}\left(21 \sigma_{5}(n)+(10-30 n) \sigma_{3}(n)-\sigma(n)\right)$

$\displaystyle f(a, b, x, y)=3 b^{2} x y^{3}+5 a b x^{3} y-3 a b x^{2} y^{2}-2 b^{2} x^{3} y$

gives

$\displaystyle 24\sum_{m=1}^{n-1} m \sigma(m) \sigma_{3}(n-m)-12\sum_{m=1}^{n-1} m \sigma_{3}(m) \sigma(n-m)$

$\displaystyle \frac{1}{10}\left(\left(6 n^{2}-5 n\right) \sigma_{3}(n)-n \sigma(n)\right)$

$\displaystyle \sum_{m=1}^{n-1} m \sigma(m) \sigma_{3}(n-m) +\sum_{m=1}^{n-1}(n-m) \sigma(m) \sigma_{3}(n-m) = n \sum_{m=1}^{n-1} \sigma(m) \sigma_{3}(n-m)$

$\displaystyle = \frac{n}{240}\left(21 \sigma_{5}(n)+(10-30 n) \sigma_{3}(n)-\sigma(n)\right)$

Adding and subtracting the last identity we get,

$\displaystyle \sum_{m=1}^{n-1} m \sigma(m) \sigma_{3}(n-m)=\frac{1}{240}\left(7 n \sigma_{5}(n)-6 n^{2} \sigma_{3}(n)-n \sigma(n)\right),$

$\displaystyle \sum_{m=1}^{n-1} m \sigma_{3}(m) \sigma(n-m)=\frac{1}{120}\left(7 n \sigma_{5}(n)+\left(5 n-12 n^{2}\right) \sigma_{3}(n)\right).$

$\displaystyle f(a, b, x, y)=a b x^{3} y-b^{2} x y^{3}$

$\displaystyle \sum_{m=1}^{n-1} m(n-m) \sigma(m) \sigma(n-m)=\frac{1}{12}\left(n^{2} \sigma_{3}(n)-n^{3} \sigma(n)\right)$

$\displaystyle f(a, b, x, y)=b^{2} y^{4}-b^{2} x y^{3}$

$\displaystyle \sum_{m=1}^{n-1} m^{2} \sigma(m) \sigma(n-m)=\frac{1}{24}\left(3 n^{2} \sigma_{3}(n)+\left(n^{2}-4 n^{3}\right) \sigma(n)\right)$

$\displaystyle f(a, b, x, y)=x y^{5}+x^{5} y-2 x^{3} y^{3}$

$\displaystyle \sum_{m=1}^{n-1} \sigma_{3}(m) \sigma_{3}(n-m)=\frac{1}{120}\left(\sigma_{7}(n)-\sigma_{3}(n)\right)$

$\displaystyle f(a, b, x, y)=x y^{5}+x^{5} y-20 x^{3} y^{3}$

$\displaystyle \sum_{m=1}^{n-1} \sigma(m) \sigma_{5}(n-m)=\frac{1}{504}\left(20 \sigma_{7}(n)+(21-42 n) \sigma_{5}(n)+\sigma(n)\right)$

Reference: https://people.math.carleton.ca/~williams/papers/pdf/249.pdf

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