I am having trouble proving this equality: $R(n) = \sigma(n/4) $ where $n/4$ is an odd positive integer or $R(n)=0$ in other case. $R(n)$ is the number of representations of $n$ as a sum of four odd squares. Let also $\sigma(n)$ be the sum of the (positive) divisors of $n\in \mathbb{Z}^{+}$. Also I do have the hint to use the following expressions \begin{align*} F(z)=\sum_{k=0}^{\infty} \frac{(2k+1)z^{2k+1}}{1-z^{4k+2}}\quad \text{and} \quad G(z)=\sum_{k=0}^{\infty} z^{(2k+1)^2} \end{align*} with the relation between the two. \begin{align*} F(z^4)=(G(z))^4. \end{align*}

As far as my attempts have gone I could only reach the following expression

\begin{align*} F(z^4)=\sum_{n=4}^{\infty} \frac{(n/4)z^n}{1-z^{2n}}\; \text{if $n$ is of the form}\; n=8k+4 \end{align*} I am interested on an expression following this calculation depending on $R$.

Another problem is knowing how to power the sum $G(z)$ as needed, should I use the Cauchy product law? Am I missing an easier way of computing it?

I have also proved

\begin{align*} n=4\left[1+\sum_{i=1}^4 (k_i^2+k_i)\right], \end{align*} therefore, $n$ must be a multiple of 4.

Any help would be deeply appreciated.

Thank you very much!


This is related to generating function, and not that hard. We want to use the relation \begin{align*} F(z^4)=(G(z))^4. \end{align*}

It is easy to see that the coefficient of $z^n$ in $(G(z))^4$ is $R(n)$. We want to show that the coefficient of $z^n$ in $F(Z^4)$ is $\sigma(n/4)$.

\begin{align*} F(z^4)=\sum_{k=0}^{\infty} \frac{(2k+1)(z^4)^{2k+1}}{1-(z^4)^{4k+2}} = \sum_{k=0}^{\infty} \sum_{j=0}^{\infty} (2k+1)(z^4)^{(2k+1)*(2j+1)} \end{align*}

So the coefficient of $z^n$ in $F(z^4)$ is the sum of $2k+1$ for which there exists a $j$ such that $n=4(2k+1)(2j+1)$, which is $\sigma(n/4).$

  • $\begingroup$ Thank you very much! Your answer is very clarifying. The last $\sigma(n) $ should be $\sigma(n/4)$, right? $\endgroup$ – Gaussicaä Feb 17 '18 at 12:21
  • $\begingroup$ Yes. Thanks. I will edit it $\endgroup$ – S. Y Feb 17 '18 at 12:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.