= Solution
The constant and first <Fourier coefficients> of $\phi^8$ are $1$ and $-16$. In the <weight-four Eisenstein basis at level two>, these force
$$
\boxed{\phi^8=\frac{16E_4(2\tau)-E_4(\tau)}{15}.}
$$
For $n>0$, its $q^n$ coefficient is $-16\sigma_3(n)+256\sigma_3(n/2)$, where the second <divisor sum> is zero if $n$ is odd. The even divisors have cube sum $8\sigma_3(n/2)$, so
$$
\sum_{d\mid n}(-1)^dd^3=16\sigma_3(n/2)-\sigma_3(n).
$$
On the other hand, expanding the eighth power of the <theta series of integer squares> counts ordered integer eight-tuples of square sum $n$. Their sign is $(-1)^{\sum_jm_j}=(-1)^n$, since $m_j^2\equiv m_j\pmod2$. If $r_8(n)$ denotes that count, then
$$
\boxed{\phi^8=\sum_{n\ge0}(-1)^nr_8(n)q^n,\qquad
r_8(n)=16(-1)^n\sum_{d\mid n}(-1)^dd^3\quad(n\ge1).}
$$
Separately \b[$r_8(0)=1$], from the all-zero tuple. The positive-divisor formula is not a formula at zero. This is the <eight-square representation formula>.
Back to article page
