The number of representations of an integer as a sum of eight triangular numbers (Q2732381)

Let NEWLINE\[NEWLINEF(z)= (1-z) \prod_{n=1}^\infty \frac {(1-q^{2n}z) (1-q^{2n}z^{-1}) (1-q^{2n})^2} {(1-q^{2n-1}z) (1-q^{2n-1}z^{-1}) (1-q^{2n-1})^2} .NEWLINE\]NEWLINE The author computes the Laurent expansion of \((F(z))^2\) in the annulus \(|q|<|z|<|q|^{-1}\), then divides both sides by \((1-z)^2\) and lets \(z\to 1\) to deduce the well-known theorem on the number of representations of an integer as a sum of eight triangular numbers. The author mentions that the analogous result for sums of four triangular numbers can be obtained using the Laurent expansion of \(F(z)\). NEWLINENEWLINENEWLINEIt may be worth pointing out that the corresponding results for sums of four and eight squares can be obtained similarly, using NEWLINE\[NEWLINEG(z)= \frac{z(1+z)} {1-z} \prod_{n=1}^\infty \frac {(1+q^{2n}z) (1+q^{2n}z^{-1}) (1-q^{2n})^2} {(1-q^{2n}z) (1-q^{2n}z^{-1}) (1-q^{2n})^2} .NEWLINE\]NEWLINE For example, computing the Laurent expansion of \((G(z))^2\) in the annulus \(|q|^2<|z|<1\), dividing by \((1+z)^2\) and letting \(z\to-1\) gives the eight squares theorem.