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This is an author-recorded argument by Plasma AI. It supplies G0 in the canonical collision sketch, not its local-factor average G1 or its smooth-input estimate G2. It is outside the independent examination of Theorems A and B. The only external asymptotic input is Mertens' product theorem, at the statement interface recorded in the analytic source interfaces.

A finite sieve inequality

Let $I$ be a real interval of length $L$, and let $f(t)=(u+bt)(v+ct)$ have integer coefficients. Suppose its number $\omega(p)$ of roots modulo each prime satisfies $0\le\omega(p)<p$ and $\omega(p)\le2$. For squarefree $d$ put $\omega(d)=\prod_{p\mid d}\omega(p)$. The Chinese remainder theorem gives

$$\#\{t\in I\cap\mathbb Z:d\mid f(t)\} =L\frac{\omega(d)}d+R_d,\qquad |R_d|\le\omega(d).$$

Changing endpoint conventions can enlarge the last constant by an absolute factor, without affecting any conclusion below. There is no dependence on $u,v,b,c$ in this error bound.

For $z\ge3$ let $\mathcal D_z$ be the squarefree positive integers $d<z$ whose prime factors have $\omega(p)>0$. Write

$$\rho(d)=\frac{\omega(d)}d,\qquad j(d)=\prod_{p\mid d}\frac{\omega(p)}{p-\omega(p)},\qquad G(z)=\sum_{d\in\mathcal D_z}j(d).$$

All three functions take value $1$ at $d=1$. Define weights supported on $\mathcal D_z$ by

$$\lambda_d=\frac{\mu(d)}{G(z)} \prod_{p\mid d}(1-\rho(p))^{-1} \sum_{\substack{k<z/d\\ k\text{ squarefree}\\(k,d)=1}} j(k),$$

where terms with $\omega(p)=0$ in $k$ have weight zero. In particular, $\lambda_1=1$. These weights satisfy the exact identity

$$\sum_{d,e\in\mathcal D_z}\lambda_d\lambda_e\rho([d,e])=\frac1{G(z)}.$$

To check it, for $r\in\mathcal D_z$ put $y_r=\sum_{r\mid d}\lambda_d\rho(d)$. Möbius inversion in the displayed definition gives $y_r=\mu(r)j(r)/G(z)$. Also

$$\rho([d,e])=\rho(d)\rho(e) \sum_{r\mid(d,e)}\frac1{j(r)},$$

as follows by checking each prime. Thus the quadratic form is $\sum_r y_r^2/j(r)=1/G(z)$.

Each factor $(1-\rho(p))^{-1}$ is at most $3$: for $p=2$ admissibility forces $\omega(2)\le1$, and for $p\ge3$ use $\omega(p)\le2$. The inner sum is at most $G(z)$, so $|\lambda_d|\le3^{\omega_0(d)}$, where $\omega_0(d)$ denotes the number of distinct prime factors of $d$. On an integer with no prime factor below $z$ dividing $f(t)$, the sum of $\lambda_d$ over $d\mid f(t)$ is $1$. Squaring and summing therefore gives

$$\#\{t\in I\cap\mathbb Z:(f(t),\prod_{p<z}p)=1\} \le\frac L{G(z)}+ \sum_{d,e\in\mathcal D_z}|\lambda_d\lambda_e|\omega([d,e]) \le\frac L{G(z)}+z^2(1+\log z)^{10}.$$

For the last inequality, the sum is at most $(\sum_{d<z}6^{\omega_0(d)})^2$. The elementary bounds $6^{\omega_0(d)}\le\tau_6(d)$ and $\sum_{d<z}\tau_6(d)\le z(1+\log z)^5$ suffice; the latter follows by summing over five of the six factors and bounding the sixth by its available length. The weights cannot be omitted from the error sum.

Uniform lower bound for the sieve denominator

Uniformly in all root-count functions above,

$$G(z)\gg\prod_{p<z}(1-\omega(p)/p)^{-1}.$$

Here is a direct proof. The elementary Chebyshev bound $\vartheta(t)=\sum_{p\le t}\log p\ll t$ follows by iterating $\prod_{N<p\le2N}p\le {2N\choose N}\le4^N$ on dyadic intervals. Partial summation gives $\sum_{p<t}(\log p)/p\le C\log t+C$ for an absolute constant $C$. Fix $a>0$ with $2Ca<1/4$ and put $w=z^a$. Let

$$W(w)=\prod_{p<w}(1+j(p))=\prod_{p<w}(1-\omega(p)/p)^{-1}.$$

Give a squarefree $d$ supported on primes below $w$ probability $j(d)/W(w)$. This is a product distribution, and its expected logarithm is

$$\sum_{p<w}\frac{j(p)}{1+j(p)}\log p =\sum_{p<w}\frac{\omega(p)\log p}{p} \le 2C\log w+2C.$$

For sufficiently large $z$ this is less than $\tfrac12\log z$. Markov's inequality shows that the total weight of $d<z$ is at least $W(w)/2$. It remains to compare $W(w)$ with the product through $z$. For $w>3$,

$$\prod_{w\le p<z}(1-\omega(p)/p)^{-1} \le\prod_{w\le p<z}(1-2/p)^{-1}\ll_a1.$$

Indeed, $(1-2/p)^{-1}=(1-1/p)^{-2}(1-1/(p-1)^2)^{-1}$; the second product converges, and Mertens' theorem bounds the first by a constant times $(\log z/\log w)^2=a^{-2}$. Bounded $z\ge3$ are absorbed by an absolute constant, since $G(z)\ge1$.

Define

$$\mathfrak S_z=\prod_{p<z}(1-\omega(p)/p)(1-1/p)^{-2}.$$

Mertens' theorem now gives the coefficient-uniform bound

$$\#\{t\in I\cap\mathbb Z:(f(t),\prod_{p<z}p)=1\} \ll \frac{L\mathfrak S_z}{\log^2z} +z^2(1+\log z)^{10}.$$

There is an absolute positive lower bound for $\mathfrak S_z$. Its factor at $2$ is at least $2$, and the product of its possible factors below $1$ is bounded below by $\prod_{p>2}(1-1/(p-1)^2)>0$. For $3\le z\le X$, the same argument gives $\mathfrak S_z\ll\mathfrak S_X$, with an absolute constant. Including a prime at an endpoint instead of using a strict cutoff changes none of these estimates.

Specialization to G0

Use the canonical sketch's notation: $n=aP\le x$, $n'=a'P'\le x$, $P,P'>x^{1/2+\eta}$, $2\le a,a'<x^{1/2-\eta}$, $g=(s(a),s(a'))$, and $0\ne h=\sigma(a')-\sigma(a)$ with $g\mid h$. The prime pairs solve

$$P=u+\beta t,\quad P'=v+\lambda t,\qquad \beta=s(a')/g,\quad\lambda=s(a)/g.$$

Enlarge their parameter interval to length $L=xg/(a s(a'))$. Uniformly in these cofactors,

$$\frac{x^{2\eta}}{1+\log x}\le L\le x(1+\log x).$$

The lower bound uses $g\ge1$ and $s(a')\le a'(1+\log a')$; the upper bound uses $g\le s(a)$ and $s(a')\ge1$.

If one linear form vanishes identically modulo a prime $p$, then $p$ divides its leading coefficient, which is at most $x^{1/2-\eta}(1+\log x)<x^{1/2+\eta}$ for large $x$. Its relevant values are primes larger than $p$, so there are no solutions. Likewise $\omega(2)=2$ permits no relevant pair. In every other case the root counts satisfy the hypotheses above.

Take $z=L^{1/4}$. For all sufficiently large $x$, every relevant prime exceeds $z$, and $\log z\gg_\eta\log x$. The sieve error $L^{1/2}(1+\log L)^{10}$ is $o(L/\log^2x)$, uniformly because of the lower bound for $L$. The positive lower bound for the singular series absorbs it into the main term. Comparing cutoffs therefore proves

$$N(a,a')\ll_\eta \frac{xg}{a s(a')\log^2x}\,\mathfrak S_x(a,a').$$

This is the pointwise G0 bound for all eligible pairs, with constants independent of the coefficients and with no separate remainder to sum. The cases $a=a'$ and $h=0$ are handled in the canonical sketch before G0. Summing this inequality still requires G1. When a prime slice has cofactors beyond the square-root range, its length can tend to zero; this proof supplies no bound for that regime.