The divisor formula for the two-square representation count

Put $g(n):={\sum }_{d\mid n}{\chi }_4(d)$. If $\gcd (m,n^{\mathrm{\prime }})=1$, every positive divisor of $m{n}^{\mathrm{\prime }}$ is uniquely of the form $ab$ with $a\mid m$ and $b\mid n^{\mathrm{\prime }}$; because odd residue classes modulo $4$ multiply and every even divisor has ${\chi }_4$-value $0$, one gets ${\chi }_4(ab)={\chi }_4(a){\chi }_4(b)$. Therefore $g(m{n}^{\mathrm{\prime }})={\sum }_{a\mid m,b\mid n^{\mathrm{\prime }}}{\chi }_4(ab)=({\sum }_{a\mid m}{\chi }_4(a))({\sum }_{b\mid n^{\mathrm{\prime }}}{\chi }_4(b))=g(m)g(n^{\mathrm{\prime }}),$ so $g$ is multiplicative.

The prime-power values of $g$ are immediate from the definition: $g(2^a)=1$ because only the divisor $1$ contributes; if $p\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}4)$, then every divisor $p^j$ is $1$ modulo $4$, so $g(p^a)=a+1$; and if $q\equiv 3(\mathrm{m}\mathrm{o}\mathrm{d}4)$, then the divisor residues alternate, so $g(q^a)=1-1+1-\cdots =\{\begin{array}{ll}1,& a\text{ even},\\ 0,& a\text{ odd}\mathrm{.}\end{array}$

By The normalized two-square count is multiplicative with the expected prime-power values, the function $r_2\mathrm{/}4$ is multiplicative with exactly the same prime-power values as in step 2.1. Hence Multiplicative functions are determined by their prime-power values gives $\frac{r_2(n)}{4}=g(n)={\sum }_{d\mid n}{\chi }_4(d)\mathrm{.}$

Among the positive divisors of $n$, the even ones contribute $0$, the divisors congruent to $1$ modulo $4$ contribute $+1$, and the divisors congruent to $3$ modulo $4$ contribute $-1$. Thus ${\sum }_{d\mid n}{\chi }_4(d)=d_1(n)-d_3(n)$, and step 3.1 gives the claimed formula for $r_2(n)$.