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This is an author-recorded argument by Plasma AI. It supports a full-slice version of the canonical sketch, in which all the large primes remain free. It does not estimate the integer solutions of the collision equation. The local estimates below therefore do not establish the smooth-input gap G2 or a solution of Problem 955.

For integers $A,B$ and $k\ge1$ put

$$F_{A,B,k}(X_1,\ldots,X_k) =A\prod_{i=1}^k(X_i+1)-B\prod_{i=1}^kX_i.$$

The class equation uses $A=\sigma(b)$ and $B=b$. For a prime $p$ and $r\ge1$, let $\rho_{p^r}(A,B;A',B')$ be

$$\frac{p^r}{\varphi(p^r)^{2k}} \#\left\{(\mathbf X,\mathbf Y)\in ((\mathbb Z/p^r\mathbb Z)^\times)^{2k}: F_{A,B,k}(\mathbf X)=F_{A',B',k}(\mathbf Y)\right\}.$$

Equality in this definition is modulo $p^r$. The normalization is the density for prime variables, with the equation imposing one congruence. All constants below may depend on the fixed integer $k$.

Uniform factors at primes not dividing the coefficients

Lemma 1. If $k\ge3$, $p\nmid ABA'B'$ and $p>k$, then, uniformly for all $r\ge1$,

$$\rho_{p^r}(A,B;A',B')=1+O_k(p^{-2}).$$

The count modulo $p$. First take $k=3$ and nonzero $a,b\in\mathbb F_p$. For a pair $(X_1,X_2)\in(\mathbb F_p^\times)^2$ put

$$U=a(X_1+1)(X_2+1),\qquad V=bX_1X_2.$$

Then $F_{a,b,3}=(U-V)X_3+U$. Write $Z$ for the number of pairs with $U=V$, $T_t$ for the number with $U=t$, and $M_t$ for the number with $U=V=t$. The number $N_3(t)$ of unit triples giving $t$ is exactly

$$N_3(t)=(p-1)^2-Z-T_t+pM_t. \tag{1}$$

Indeed, a nonzero coefficient $U-V$ allows one value of $X_3$ unless $t=U$, in which case that value is zero and is excluded. A zero coefficient allows all $p-1$ unit values exactly when $t=U$.

We have $Z\le p-1$. To see this, fix $X_1$ and solve the linear equation $(a(X_1+1)-bX_1)X_2=-a(X_1+1)$; its coefficient and right side cannot both vanish, because $b\ne0$. Also $T_0=2p-3$, while $T_t\le p-1$ for $t\ne0$. Finally $M_0=0$, and $M_t\le2$ for $t\ne0$: the product $X_1X_2=t/b$ and sum $X_1+X_2=t/a-t/b-1$ determine the two variables as roots of a quadratic. Formula (1) consequently gives, uniformly in $t$,

$$\Pr(F_{a,b,3}=t)=\frac1p+O(p^{-2}). \tag{2}$$

For $k>3$, fix the first $k-3$ unit variables. Unless one is $-1$, the remaining three variables have two nonzero coefficients, and (2) applies. If one is $-1$, the polynomial is a nonzero constant times the product of the last three variables, which is uniform on $\mathbb F_p^\times$. The latter case has probability at most $(k-3)/(p-1)$. Thus (2) holds with $O_k(p^{-2})$ for every $k\ge3$.

Write the two value distributions as $1/p+e_t$ and $1/p+e'_t$. Both errors sum to zero and have absolute value $O_k(p^{-2})$. Their collision probability is therefore

$$\frac1p+\sum_t e_te'_t=\frac1p+O_k(p^{-3}),$$

which proves the assertion for $r=1$.

Lifting to prime powers. Put $G=F_{A,B,k}(\mathbf X)- F_{A',B',k}(\mathbf Y)$ and $N=2k$. At a critical point of one block, no coordinate can equal $-1$: a derivative in another coordinate would then have its second, nonzero product term only. With every coordinate different from $0,-1$, write $m=F_{A,B,k}(\mathbf X)$ and $n=B\prod X_i$. Direct differentiation gives

$$\frac{\partial F}{\partial X_i} =\frac{X_i m-n}{X_i(X_i+1)}.$$

If all derivatives vanish, $m\ne0$ and all the coordinates equal a single $t$. That $t$ satisfies $A(t+1)^{k-1}=Bt^{k-1}$, a nonzero polynomial equation of degree at most $k-1$. Hence $G$ has at most $(k-1)^2$ critical points in the unit domain modulo $p$.

At such a point the Hessian of one block has zero diagonal and every off-diagonal entry equal to $c=-Bt^{k-2}/(t+1)\ne0$. It is $c(J-I)$ and has determinant $(k-1)c(-c)^{k-1}$. Thus the Hessian of $G$ is invertible when $p>k$.

Every nonsingular zero modulo $p$ has exactly $p^{(N-1)(r-1)}$ lifts modulo $p^r$, so its contribution to $\rho_{p^r}$ is independent of $r$. A critical residue class contributes $O_k(p^{2-N})$, uniformly in $r$. Here is an elementary bound for that last assertion. In a critical class write the variables as $\mathbf z+p\mathbf u$. If $G(\mathbf z)$ is not divisible by $p^2$, there are no lifts past modulus $p$. Otherwise, after division by $p^2$, the resulting polynomial modulo $p$ has an invertible quadratic part. Such a polynomial has at most $2p^{N-1}$ zeros and at most one critical point. The first bound follows by diagonalizing its quadratic part and solving a genuine quadratic in one coordinate; the second follows from its invertible Hessian.

If $B_r$ is the normalized root count for any polynomial with this quadratic reduction, its nonsingular classes contribute at most $2$. Its possible critical class, when it lifts, rescales in the same way, giving

$$B_r\le2+p^{2-N}B_{r-2},\qquad B_0=1,\quad B_1\le2.$$

Thus $B_r\le2/(1-p^{2-N})$. The original critical class has Haar measure $p^{-N}$ and its equation acquired a factor $p^2$, giving the asserted $O_k(p^{2-N})$ after conditioning on units. This also bounds the difference between its contribution at modulus $p$ and at modulus $p^r$. There are at most $(k-1)^2$ such classes, and $N\ge6$, so the estimate modulo $p$ proves the lemma for every $r$. $\square$

The factors in Lemma 1 are summable over their primes. This is a statement about the full equation with at least three variables on each side. It does not assert summability for the two-linear-form factors of G1.

A bound at every primitive coefficient pair

Lemma 2. Suppose $p\nmid(A,B)$ and $p\nmid(A',B')$, with $k\ge2$. Uniformly for all $r\ge1$,

$$\rho_{p^r}(A,B;A',B')\le1+O_k(p^{-1}).$$

Proof. For $j\ge1$, and $h\ge1$ independent unit variables, we first have the uniform estimate

$$\Pr\bigl(p^j\mid F_{A,B,h}\bigr) \ll_h (j+1)^{h-1}p^{-j}. \tag{3}$$

If $p\mid A$ then $B$ is a unit and the probability is zero. Otherwise the substitution $R_i=1+1/X_i$ is a measure-preserving bijection from units modulo $p^j$ to residues different from $1$ modulo $p$. It changes the congruence to $\prod R_i=B/A$. For unrestricted uniform residues, the probability of a specified product is $O_h((j+1)^{h-1}p^{-j})$: for a nonzero target of valuation $v<j$, sum over the $\binom{v+h-1}{h-1}$ valuation tuples and use uniformity of the unit product; for the zero target, sum the geometric tail of valuation tuples whose total is at least $j$. Conditioning the $R_i$ to avoid $1$ modulo $p$ costs at most $(1-1/p)^{-h}\le2^h$. This proves (3).

Fix the first $k-1$ variables of each block. The equation in the two remaining unit variables is $CX-DY=H$, with $C=F_{A,B,k-1}$ and $D=F_{A',B',k-1}$. Set $v=\min(v_p(C),v_p(D),r)$. Its conditional probability is at most $p^{v-r}p/(p-1)$, including $v=r$. The two prefixes are independent, so (3) gives

$$\begin{aligned} \rho_{p^r} &\le\frac p{p-1}\,\mathbb E p^v\\ &\le\frac p{p-1} \left(1+O_k\left(\sum_{j\ge1}(j+1)^{2k-4}p^{-j}\right)\right) =1+O_k(p^{-1}). \end{aligned}$$

The constants do not depend on the valuations of the coefficients. $\square$

Averaging the local majorant over small cofactors

Let $y\to\infty$, $L=\log\log y$ and $W=L/(4\log L)$. Call $b<y$ locally good if

$$v_p(\sigma(b))>v_p(b)\quad(p\le W),\qquad p\nmid(b,\sigma(b))\quad(p>W).$$

The harmonic estimates in the proofs of conditions 2 and 3 in the good-cofactor calculation apply directly to $b$. They show that the other $b<y$ have reciprocal sum $o(\log y)$: the proof of condition 2 bounds all violating $c<y$ before inserting its outer prime. Neither estimate requires two large prime factors of $b$.

For locally good $b,b'$ set $w=W(b)=\prod_{p\le W}p^{v_p(b)}$. If $W(b)\ne W(b')$, define $\mathfrak U_k(b,b')=0$. Otherwise define

$$\mathfrak U_k(b,b')= \prod_{p\le W}\sup_{r\ge1}\rho_{p^r}(\sigma(b),b;\sigma(b'),b') \prod_{p>W}\max\left\{1,\sup_{r\ge1} \rho_{p^r}(\sigma(b),b;\sigma(b'),b')\right\}.$$

This is a local majorant, not an asserted asymptotic singular series; the maximum at the larger primes discards factors smaller than one. For fixed $b,b'$ the product converges: Lemma 1 applies outside finitely many coefficient primes. If $W(b)\ne W(b')$, the two integer outputs cannot agree when all prime variables exceed $W$, since their valuations at a small prime are respectively $v_p(b)$ and $v_p(b')$.

Proposition 3. For each fixed $k\ge3$,

$$\sum_{\substack{b,b'<y\\b,b'\ \mathrm{locally\ good}}} \frac{\mathfrak U_k(b,b')}{bb'}\ll_k\log^2y. \tag{4}$$

Small primes. If $p\le W$ and $v_p(b)=v_p(b')=e$, divide both polynomials by $p^e$. Their first coefficients are divisible by $p$ and their second coefficients are units. Fixing all but one variable shows that each divided value is uniform on the units modulo every $p^j$: the remaining linear coefficient is a unit and its intercept is divisible by $p$. Therefore

$$\sup_{r\ge1}\rho_{p^r}=p^e\frac p{p-1}. \tag{5}$$

For $r\le e$ the factor is $p^r$, and for $r>e$ it is the expression in (5).

Larger primes. For large $y$ we have $W>k$. Lemma 2 applies at every $p>W$ by local goodness, and Lemma 1 applies unless $p\mid b\sigma(b)b'\sigma(b')$. Hence for a constant $C_k$ and

$$G_k(n)=\prod_{\substack{p>W\\p\mid n\sigma(n)}}(1+C_k/p),$$

we have

$$\mathfrak U_k(b,b') \ll_k w\prod_{p\le W}\frac p{p-1}\,G_k(b)G_k(b'). \tag{6}$$

The harmonic mean of $G_k$. Write $S_W=\prod_{p\le W}(1-1/p)$. For $w$ supported on primes up to $W$, the needed bounds are

$$\sum_{\substack{b<y\\W(b)=w}}\frac{G_k(b)}b \ll_k\frac{G_k(w)}w\,S_W\log y, \qquad \sum_{w}\frac{G_k(w)^2}w\le(1+o(1))S_W^{-1}. \tag{7}$$

Here are details of the only distribution input in (7). The function $G_k$ is submultiplicative on coprime arguments, so an Euler product bounds its harmonic sum. For a prime $q>W$, expand $\prod_{p\mid q+1,\ p>W}(1+C_k/p)=\sum_{d\mid q+1}\xi(d)$, where $\xi$ is squarefree-supported and $\xi(p)=C_k/p$. For $1<d\le\sqrt y$, partial summation of Brun–Titchmarsh gives

$$\sum_{\substack{q<y\ \mathrm{prime}\\q\equiv-1\pmod d}}\frac1q \le\frac{2\log\log y+O(1)}{\varphi(d)}.$$

The moduli here are odd and exceed $W>3$; their first possible prime in the class is at least $2d-1$, so the stated interval condition causes only a boundary term of order $1/d$. For $d>\sqrt y$, the bound $2(1+\log y)/d$ for all integers in the class suffices. Consequently

$$\sum_{q<y\ \mathrm{prime}}\frac{G_k(q)-1}{q} \ll_k L\sum_{d>1}\frac{\xi(d)}{\varphi(d)}+o(1) \ll_k\frac{L}{W\log W}+o(1)=O_k(1). \tag{8}$$

For the large-$d$ term use $\sum_{d>\sqrt y}\xi(d)/d\le y^{-1/4}\sum_d\xi(d)/\sqrt d=O_k(y^{-1/4})$. The extra factor $1+C_k/q$ from $q\mid n$ contributes $o(1)$ in (8). For prime powers $q^j$ with $j\ge2$, the bound $\sigma(q^j)\le2q^j$ gives $G_k(q^j)\ll_k q^{j\epsilon_W}$ with $\epsilon_W=O_k(1/(W\log W))$; their Euler terms for $q>W$ sum to $o(1)$. Mertens' product theorem and (8) prove the first part of (7).

For the second part, at a prime $q\le W$ the factors of $G_k(q^j)$ come only from $\sigma(q^j)$. Their logarithm is at most $C_k(j+1)\log q/(W\log W)$. Comparing Euler factors with $(1-1/q)^{-1}$, their total relative excess is $O_k((W\log W)^{-1}\sum_{q\le W}(\log q)/q)=O_k(1/W)$. This proves the second part of (7), also with the square on $G_k$.

Finally sum (6), first over cofactors with the same $w$ and then over $w$. By (7) the result is bounded by

$$O_k(\log^2y)\,S_W\sum_w\frac{G_k(w)^2}w \ll_k\log^2y,$$

which proves (4). $\square$

Relation to the whole target

Theorem C of the canonical sketch writes almost all retained inputs as $n=bP_1\cdots P_k$, with $b\le x^\tau$ and $P_i>x^\theta$. For fixed $k,\theta$, the number of such inputs with a specified $b$ is $O_{k,\theta}(x/(b\log x))$: count the largest prime by the elementary upper bound for $\pi$, and sum $1/(P_2\cdots P_k)$ for the others. Mertens' theorem bounds each reciprocal prime sum by $\log(1/\theta)+o(1)$. Thus the non-locally-good $b$ cost $o(x)$ inputs.

One may also discard every fixed finite list of small $k$ while losing an arbitrarily small input proportion as $\theta\to0$. To see this, fix $0<\beta<1/2$ and count the primes in $(x^\theta,x^\beta]$ dividing a uniform integer at most $x$. Its mean is $M=\log(\beta/\theta)+o(1)$ and its variance is at most $M+o(1)$. Expanding the first two moments proves these statements: the total floor errors are $O(\pi(x^\beta)^2/x)=o(1)$, and Mertens' theorem gives the mean. For any fixed $K<M$ Chebyshev therefore bounds the proportion with at most $K$ such primes by

$$\frac{M+o(1)}{(M-K)^2}.$$

This tends to zero when $\theta\to0$. In particular a whole-target route using only the classes $k\ge3$ is sufficient; it need not settle G1 or the quadratic slice first.

Proposition 3 supplies the local-factor average for that route. The continuous prime-density integral also has the required size. For fixed $b$ with $s(b)\ge\delta b/2$, solve for $P_1$ while the other variables are held fixed. On the positive real domain the derivative is $\sigma(b)\prod_{i>1}(P_i+1)-b\prod_{i>1}P_i \ge s(b)\prod_{i>1}P_i$, and $\log P_1\ge\theta\log x$. Integrating the other variables with measure $dP_i/(P_i\log P_i)$ bounds the output density by $O_{\delta,\theta,k}(1/(b\log x))$. On an output interval $[0,Cx]$ the overlap of two such densities is at most $O_{C,\delta,\theta,k}(x/(bb'\log^2x))$. Equation (4) then bounds the sum of these local majorants times continuous overlaps by $O(x)$.

Remaining global obligation. Relate these local and continuous counts to the off-diagonal prime solutions of $F_{\sigma(b),b,k}(\mathbf P)=F_{\sigma(b'),b',k}(\mathbf P')$ uniformly over the growing small cofactors and the prime boxes, with an aggregate error $O(x)$. Equal input tuples contribute at most $x$ and can be separated. Neither a bound for every residue modulus nor the continuous integral controls the integer discrepancy. In particular, applying a separate bound with a unit error to each short slice still incurs the loss in Proposition D of the canonical sketch. No estimate for that discrepancy is proved here.

Increasing $k$ does not make the leading homogeneous form nonsingular. It is $s(b)\prod X_i-s(b')\prod Y_i$; all its first derivatives vanish exactly when at least two coordinates of each block vanish. That singular locus has codimension four for every $k\ge2$. Lemma 1 concerns the affine polynomial and its unit residue classes, and supplies no bound for the contribution of short boxes or for oscillatory sums over prime variables. A global theorem used here must allow this leading form and coefficients growing with $x$; none has been established in this argument.

With such a global estimate, the negligible cofactor removal, the small-$k$ deletion, Theorem C's decomposition and Theorem B's output truncation would give the full density-zero preimage conclusion. Without it, Lemmas 1 and 2 and Proposition 3 are local statements only. They are author-recorded, have not undergone independent review, and do not certify G1-core or G2.