As is well known due to Lagrange the number of ways to write $n$ as a sum of four squares is given by $$ r_4(n)=8\sum_{d|n} d $$ if $n$ is odd.

Now define $$ \tilde{r}_4(n)=\#\left\{1 + \sum_{i=1}^{4} n_i(n_i+1) = n \,\Big|\, (n_1,n_2,n_3,n_4) \in {\mathbb Z}^4 \right\} $$ and clearly only for odd $n$ is $\tilde{r}_4(n)>0$ since $n_i(n_i+1)$ is even.

I'm trying to show that $$\tilde{r}_4(n)=2\,r_4(n)$$ for odd $n$, but I lack a clever idea :-(

Background: This problem arises if one tries to proof $$\theta_2(0;q)^4 - \theta_3(0;q)^4 + \theta_4(0;q)^4 = 0$$ where $\theta_i(0;q)$ is the Jacobi Theta function. So maybe there is another way to proof this identity from which the original question then follows.


1 Answer 1


$1+\sum_{k=1}^{4}n_k(n_k+1)=n$ is equivalent to

$$\sum_{k=1}^{4}(2n_k+1)^2 = 4n $$ so $\tilde{r}_4(n)$ accounts for the number of ways for writing $4n$ as the sum of four odd squares.
On the other hand $a^2+b^2+c^2+d^2 = 4n$ implies that $a,b,c,d$ have the same parity, hence $$\tilde{r}_4(n) = r_4(4n)-r_4(n) =2\,r_4(n).$$

  • $\begingroup$ Nice elegant way to solve it. Thanks so much! $\endgroup$
    – Diger
    Oct 9, 2018 at 17:04
  • $\begingroup$ +1 for simplest proof of the theta function identity. $\endgroup$
    – Paramanand Singh
    Oct 28, 2018 at 4:08

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