By the orthogonality of characters, we have

where

Then Cauchy-Schwarz gives

By Parseval’s identity and the divisor bound, we have

for any \(\varepsilon {\gt}0\). Using Cauchy-Schwarz again, for any \(i\le d\) we have

where

Let \(ir\) be the largest multiple of \(i\) in \([1,d]\). Then

where \(\tilde N\) is the number of

such that

for all \(j\leq d\), and

Let us write

where \(\tilde{N}_1\) is the contribution to \(\tilde N\) from tuples with \(\prod _{j\equiv 0\bmod i}z_j^j= \prod _{j \equiv 0\bmod i}z_j'^j\), and \(\tilde{N}_2\) is the contribution of the complementary tuples.

Then by the divisor bound we have

for any \(\varepsilon {\gt}0\). In order to bound \(\tilde{N}_2\), we first note that \(a-b\mid a^i-b^i\) for any integers \(a\neq b\) and \(i\geq 1\). Thus for any integers \(n\neq 0\) and \(i\geq 2\),

This follows from the divisor bound and the fact that \((x+d)^i-x^i-n\) is a polynomial of degree \(i-1\). (Importantly, this argument fails when \(i=1\), since then the polynomial \((x+n)^i-x^i-n\) is identically \(0\).) Hence, on appealing to the divisor bound, we obtain

Combining 3.1.5, 3.1.9, 3.1.10 and 3.1.11, we deduce that

Plugging 3.1.2 and 3.1.16 back into 3.1.1, we conclude that

which is the desired bound.

Take any sets \(I,I',I''\subset [d]\). Let \((x_1,\ldots , x_d,y_1,\ldots , y_d,z_1,\ldots , z_d)\) be a tuple counted by \(B_d(\mathbf{c},\mathbf{X},\mathbf{Y},\mathbf{Z})\). We fix a choice of \( x_i,y_i,z_i \in \mathbb {Z} \) for all indices \(i\) in \(I,I',I''\), respectively, and define

Then \(\gcd (a_1,a_2,a_3)=1\) and \(\gcd (x,y,z)=1\). According to Heath-Brown  [ 6 , Lemma 3 ] , the number of triples \((x,y,z)\) that contribute to \(B_d(\mathbf{c},\mathbf{X},\mathbf{Y},\mathbf{Z})\) is

Moreover, by the divisor bound, any triple \((x,y,z)\) corresponds to \(O_{\varepsilon }(\Delta ^{\varepsilon })\) choices of \(x_i,y_j,z_r\) with \(i\not\in I, j\not\in I', r\not\in I''\). We arrive at the desired upper bound by summing over the choices \(x_i,y_i,z_i\in \mathbb {Z}\), with \(i\) in \(I,I',I''\).

We may assume a factorisation \(g(x)=l_1(x)^{e_1}\dots l_t(x)^{e_t}\), with pairwise non-proportional linear polynomials \(l_1,\dots ,l_t\in \mathbb {C}[x]\) and exponents \(e_1,\dots ,e_t\in \mathbb {N}\) such that \(e_1=1\). But \({\mathbb {C}}[x]\) is a unique factorisation domain and so we can apply Eisenstein’s criterion with the prime \(l_1\) in order to deduce that \(g(x)-y^r\) is irreducible over \( {\mathbb {C}}[y]\). It then follows that \(g(x)-y^r\) is irreducible over \( \mathbb {C}\), as claimed in the lemma.

Suppose \(n=a_1x_1^2+a_2x_2^2\). Then \(a_1n=a_1^2x_1^2+a_1a_2x_2^2\). Let \(D\) be the squarefree part of \(a_1a_2\). Then the number of solutions \((x_1,x_2)\) with \(|x_i|\leq X_i\) to this equations is at most the number of solutions \((m_1,m_2)\) to \(a_1n=m_1^2+Dm_2^2\) with \(|m_i|\leq X_ia_1a_2\). For any such solution, we have \(m_1+\sqrt{-D}m_2\mid a_1n\) in \(\mathbb {Q}(\sqrt{-D})\). The claim follows from the divisor bound in quadratic fields, in Lemma 3.8.

Mimics the proof over the integers using the fundamental theorem of arithmetic, but with ideals [See link in comment]

We fix a choice of non-zero integer \(z\in [-Z,Z]\), of which there are \(O(Z)\). When \(z\) is fixed, the resulting equation defines a curve in \(\mathbb {A}^2\) and we can hope to apply work of Bombieri-Pila  [ 2 , Theorem 4 ] , which would show that the equation has \(O_{\varepsilon ,D}(\max (X,Y)^{\frac{1}{\max (p,q)}+\varepsilon })\) integer solutions in the region \(|x|\leq X\) and \(|y|\leq Y\), where the implied constant only depends on \(\varepsilon \) and \(D\). This is valid only when the curve is absolutely irreducible, which we claim is true when \(z\neq 0\). But, for fixed \(z\in \mathbb {Z}_{\neq 0}\) the polynomial \(a_2y^q+a_3z^r\) has non-zero discriminant as a polynomial in \(y\). Hence the claim follows from Lemma ??. Rather than appealing to Bombieri-Pila, however, we can get a sharper bound by using work of Heath-Brown  [ 8 , Theorem 15 ] . For fixed \(z\in \mathbb {Z}_{\neq 0}\) this gives the bound \(O_{\varepsilon ,D}(\min (X^{\frac{1}{q}},Y^{\frac{1}{p}})(XY)^\varepsilon )\) for the number of available \(x,y\).

Suppose now that \(p=q\geq 2\). Then, for given \(z\in \mathbb {Z}_{\neq 0}\), we are left with counting the number of integer solutions to the equation \(N=a_1x^p+a_2y^p\), with \(|x|,|y|\leq \max (X,Y)\), and where \(N=-a_3z^r\). For \(p=2\) this is a classical problem in quadratic forms. The bound \(O_{\varepsilon ,D}((|a_1a_2N|XY)^\varepsilon )\) follows from Heath-Brown  [ 7 , Theorem 3 ] , for example. For \(p\geq 3\) we obtain a Thue equation. According to work of Bombieri and Schmidt  [ 3 ] , there are at most \(O(p^{1+\omega (|N|)})\) solutions, for an absolute implied constant. Using the bound \(\omega (|N|)\ll \log (3|N|)/(\log \log (3|N|))\), this is \(O_{\varepsilon ,D}((|a_3|Z)^\varepsilon )\), which thereby completes the proof of the lemma.

Let \((x_1,\ldots , x_d,y_1,\ldots , y_d,z_1,\ldots , z_d)\) be a tuple counted by \(B_d(\mathbf{c},\mathbf{X},\mathbf{Y},\mathbf{Z})\). For any integers \(p,q\geq 1\), we fix all but \(x_p\) and \(y_q\) and apply the first part of Lemma 3.12. This gives

from which the statement of the lemma easily follows.

Let \((x_1,\ldots , x_d,y_1,\ldots , y_d,z_1,\ldots , z_d)\) be a tuple counted by \(B_d(\mathbf{c},\mathbf{X},\mathbf{Y},\mathbf{Z})\). We collect together all indices which are multiples of \(p\) for any integer \(p\geq 2\). Applying the second part of Lemma 3.13 now gives

from which the statement easily follows.