 Solution

The constant and first Fourier coefficients of ${\varphi }^8$ are $1$ and $-16$. In the weight-four Eisenstein basis at level two, these forceFor $n>0$, 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 $8{\sigma }_3(n/2)$, soOn 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 $(-1{)}^{{\sum }_j{m}_j}=(-1{)}^n$, since $m_j^2\equiv m_j(\mathrm{mod}2)$. If $r_8(n)$ denotes that count, thenSeparately $r_8(0)=1$, from the all-zero tuple. The positive-divisor formula is not a formula at zero. This is the eight-square representation formula.