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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!

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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 $\sigma(n/4).$

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  • $\begingroup$ 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

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