Since \(D<0\), the equation \(t^2+|D|u^2=4\) forces \(|u|\le 1\). The solutions with \(u=0\) are \((t,u)=(\pm 2,0)\). When \(|u|=1\), \(t^2=4-|D|\), so \(t=\pm 1\) when \(|D|=3\) and \(t=0\) when \(|D|=4\). Thus \(|D|=3\) gives 6 solutions, \(|D|=4\) gives 4, and every larger value gives two. These are the three values of \(w_D\).

The matrix of \(Q\) has determinant \(-D/4\) up to the usual factor \(4\). Explicitly, \[\det \begin {pmatrix}2a&b\\b&2c\end {pmatrix}=-D.\] An element \(\gamma \in \mathrm {SL}_2(\mathbb Z)\) replaces that matrix by a congruent matrix, so it preserves \(D\). Let \(\tau =(-b+\sqrt {D})/(2a)\) be the root in the upper half plane, taking the square root of positive imaginary part. The factorization \(Q(x,y)=a(x-\tau y)(x-\overline {\tau } y)\) shows that an automorph permutes the two roots. Determinant \(1\) preserves the upper half plane, so the automorph fixes \(\tau \). If \(\gamma =\begin {pmatrix}p&q\\r&s\end {pmatrix}\), the fixed-point equation is \(r\tau ^2+(s-p)\tau -q=0\). The minimal equation \(a\tau ^2+b\tau +c=0\) is irreducible over \(\mathbb Q\), so there is a rational scalar \(u\) with \(r=au\), \(s-p=bu\), and \(q=-cu\). First, \(\gcd (a,b,c)=1\). A prime dividing all three coefficients square-divides \(D\). The odd part of a fundamental discriminant is squarefree, so that prime is not odd. If the prime is \(2\), then \(D=4m\). Writing the coefficients as twice integers \(a'\), \(b'\), and \(c'\) gives \(m=b'^2-4a'c'\equiv b'^2\pmod 4\), and a square modulo \(4\) is \(0\) or \(1\), contrary to the fundamental condition on \(m\). Bézout's identity therefore writes \(1\) as an integer combination of \(a\), \(b\), and \(c\), and the rational scalar \(u\) is an integer. Set \(t=p+s\). The determinant \(ps-qr=1\) becomes \((t^2-Du^2)/4=1\). Conversely each solution \((t,u)\) determines \[p=\frac {t-bu}{2},\qquad s=\frac {t+bu}{2},\qquad q=-cu,\qquad r=au.\] The same parity of \(t\) and \(bu\) makes \(p\) and \(s\) integers, since \(t^2\equiv Du^2\pmod 2\) and \(D\equiv b^2\pmod 2\). The form produced by \(\gamma \) has the same root and the same discriminant, so it is a positive multiple of \(Q\). The discriminant scales by the square of that multiple, and the multiple is positive, so the multiple is \(1\). Thus these \(w_D\) matrices are exactly the automorphs of every form of discriminant \(D\).

Let \(R(n)\) be the number of pairs \(\bigl (Q,(x,y)\bigr )\) with \(Q\in \mathcal F\) and \(Q(x,y)=n\). Let \(R^\ast (n)\) count only the pairs with \(\gcd (x,y)=1\). A nonzero vector has content \(d=\gcd (x,y)\), and \(Q(dv)=d^2Q(v)\), so \[R(n)=\sum _{d^2\mid n}R^\ast (n/d^2).\] The stabilizer in the automorph group of any primitive vector is trivial: an automorph fixing a vector has eigenvalue \(1\), hence trace \(2\) because the determinant is \(1\), hence \(t=2\) and \(u=0\). The group therefore acts freely on the proper representations of each form. It remains to show that the orbits are indexed by the residues \(b\) modulo \(2n\) with \(b^2\equiv D\pmod {4n}\).

Given a proper representation of \(n\) by a vector \(v\), complete \(v\) to a basis \((v,w)\) of determinant \(1\). In that basis the form is \(nx^2+Bxy+Cy^2\) with \(B^2-4nC=D\). Replacing \(w\) by \(w+tv\) sends \(xv+yw\) to \((x+ty)v+yw\). The term \(n(x+ty)^2\) contributes the extra middle term \(2nt\,xy\), so the middle coefficient changes by \(2nt\) and \(B\) modulo \(2n\) is invariant. An automorph changes the vector inside one orbit and does not change this residue. In the other direction, a residue \(b\) produces the positive definite form \(nx^2+bxy+cy^2\) with \(c=(b^2-D)/(4n)\). That form has discriminant \(D\), so it is equivalent to exactly one representative in \(\mathcal F\). Transporting \((1,0)\) by such an equivalence yields a proper representation with that residue. If two proper representations have the same residue \(b\), replace each second basis vector by a shear so that the middle coefficient equals \(b\). The discriminant identity then forces both coordinate forms to equal \(nx^2+bxy+cy^2\) with the same \(c=(b^2-D)/(4n)\). Those two determinant-\(1\) bases therefore belong to the same class representative. If \(M_1\) and \(M_2\) are the basis matrices, \(M_1M_2^{-1}\) carries one represented vector to the other and preserves the form, since both bases pull it back to that coordinate polynomial. The number of such residues is therefore the number of orbits, and \(R^\ast (n)=w_D\rho (n)\), where \(\rho (n)\) denotes that number of residues.

The count \(\rho \) is multiplicative on coprime arguments. If \(n_2\) is odd and coprime to \(n_1\), a residue modulo \(2n_1n_2\) is a pair of a residue modulo \(2n_1\) and a residue modulo \(n_2\). The square condition splits, and for odd \(n\) the count \(\rho (n)\) equals the number of solutions of \(v^2\equiv D\pmod n\). Indeed \(\gcd (4,n)=1\), so the condition modulo \(4n\) splits into conditions modulo \(4\) and modulo \(n\), and exactly one of the two lifts modulo \(2n\) has the parity required by \(D\) modulo \(4\). Thus \(\rho (n_1n_2)=\rho (n_1)\rho (n_2)\).

For an odd prime \(p\) the same count is the number of roots of \(v^2-D\) modulo \(p^k\). If \(p\) does not divide \(D\) and \(\chi _D(p)=1\), there are two simple roots modulo \(p\), and the derivative \(2v\) is invertible at each, so each lifts uniquely and \(\rho (p^k)=2\). If \(\chi _D(p)=-1\), there is no root and \(\rho (p^k)=0\). If \(p\) divides \(D\), then \(p\) divides \(D\) exactly once. Modulo \(p\) the only root is \(0\), so \(\rho (p)=1\). Modulo \(p^k\) with \(k\ge 2\) the valuation \(v_p(D)=1\) is odd, while a square has even valuation, so \(\rho (p^k)=0\).

At the prime \(2\), first take \(n=1\): exactly one of \(b=0,1\) satisfies \(b^2\equiv D\pmod 4\), so \(\rho (1)=1\). If \(D\) is odd and \(D\equiv 5\pmod 8\), then for \(a\ge 1\) the modulus \(2^{a+2}\) is a multiple of \(8\), and no square is congruent to \(5\) modulo \(8\), so \(\rho (2^a)=0\). If \(D\equiv 1\pmod 8\), the two odd residues modulo \(4\) satisfy the congruence modulo \(8\). A solution modulo \(2^{a+1}\) lifts to a solution modulo \(2^{a+2}\) by \(b=b_0+t2^{a+1}\). The cross term determines \(t\) uniquely because \(b_0\) is odd, and the quadratic term in \(t\) is divisible by \(2^{a+3}\). Thus \(\rho (2^a)=2\) for every \(a\ge 1\). If \(D\) is even, inspect \(b\) modulo \(4\) against the condition modulo \(8\). When \(D\equiv 4\pmod 8\) the only solution is \(b\equiv 2\pmod 4\). When \(D\equiv 0\pmod 8\) the only solution is \(b\equiv 0\pmod 4\). Thus \(\rho (2)=1\). For \(a\ge 2\) write \(D=4m\). If \(m\equiv 2\pmod 4\), then \(v_2(D)=3\). A square cannot have that valuation, and \(3\) is strictly smaller than the modulus \(2^{a+2}\), so there is no solution. If \(m\) is odd, a solution would have valuation exactly \(2\), hence the shape \(b=2c\) with \(c\) odd, and then \(c^2\equiv m\pmod 4\). An odd square is \(1\) modulo \(4\), while \(m\equiv 3\pmod 4\). There is again no solution, so \(\rho (2^a)=0\) for \(a\ge 2\).

Let \(\psi (n)=\sum _{d\mid n}\chi _D(d)\) and \(g(n)=\sum _{d^2\mid n}\rho (n/d^2)\). Both functions are multiplicative, since a square divisor of a product of coprime arguments splits, so it is enough to compare them on prime powers. If \(\chi _D(p)=1\), then \(\rho (p^j)=2\) for \(j\ge 1\). For even \(k=2m\) the sum \(g(p^k)\) is \(m\) terms equal to \(2\) and one term \(\rho (1)\), hence \(k+1\). For odd \(k=2m+1\) it is \(m+1\) terms equal to \(2\), hence again \(k+1\). The character sum is the same geometric sum. If \(\chi _D(p)=-1\), then \(\rho (p^j)=0\) for \(j\ge 1\), so \(g(p^k)\) keeps only \(\rho (1)\) when \(k\) is even and is \(0\) when \(k\) is odd. The alternating character sum has those same two values. If \(\chi _D(p)=0\), then \(\rho (p)=1\) and \(\rho (p^j)=0\) for \(j\ge 2\), so every \(g(p^k)\) equals \(1\), as does \(\psi (p^k)\). Therefore \(g=\psi \), and \[R(n)=w_D\sum _{d\mid n}\chi _D(d).\]

Absolute convergence for \(\operatorname {Re}s>1\) lets the zeta sums be rearranged. The form has no real root and \(a>0\), so its minimum \(\mu \) on the unit circle is positive and \(Q(x,y)\ge \mu (x^2+y^2)\). The comparison series converges for \(\operatorname {Re}s>1\) because, in polar coordinates, the tail is a constant multiple of \(\int _1^\infty r^{1-2s}\,dr\). Hence \[\sum _{Q\in \mathcal F}E_Q(s) =\sum _{n\ge 1}R(n)\,n^{-s}.\] The representation count and the substitution \(n=dm\) give \[w_D\sum _{d,m\ge 1}\chi _D(d)\,(dm)^{-s} =w_D\,\zeta (s)\sum _{d\ge 1}\chi _D(d)\,d^{-s}.\] Complete multiplicativity expands each Euler factor \((1-\chi _D(p)p^{-s})^{-1}\) as a geometric series. The product in the center note is therefore the Dirichlet series, which is the identity claimed above.

For the illustration \(D=-4\), repeat the shear \(\begin {pmatrix}1&k\\0&1\end {pmatrix}\), which replaces \(b\) by \(b+2ak\) and does not change \(a\), until \(|b|\le a\). Whenever \(a>c\), apply \(\begin {pmatrix}0&-1\\1&0\end {pmatrix}\). That matrix exchanges the outer coefficients and changes the sign of the middle coefficient, so it replaces \(a\) by the smaller positive integer \(c\). Each such swap decreases \(a\), and the process stops at a representative with \(|b|\le a\le c\). Then \(b\) is even, \(ac=(b/2)^2+1\), and \(a^2\le ac\le a^2/4+1\), so \(a=1\), \(b=0\), and \(c=1\). The only class is \(x^2+y^2\), and \(w_{-4}=4\).