Independent record: Erdős Problem 940

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The companion working files this record names (positive_density_reduction.md and the others listed at its end) were not retained; the names are kept so the citations stay readable, and the retained re-derivations are the topical notes of this folder.

1. Scope and outcome

This investigation used only problem page and the supplied papers. No web search was made, and no earlier research notes were assumed. Assertions in digests about unspecified “current research” were not treated as results. The full supplied Ghidelli paper and Voight chapter were read by the collaborators using their results. The full Beyer de Ryke paper was also read for its canonical decomposition and counting framework, not to re-audit its known cubic theorem.

Write

$$\mathcal P_r=\{m\geq1:p\mid m\Longrightarrow p^r\mid m\},\qquad S_r=\bigcup_{j=1}^r j\mathcal P_r.$$

The supplied background establishes $\underline d(\mathbb N\setminus S_3)>0$. It does not establish $d(S_3)=0$.

The main argument developed here is a quartic complement argument:

Quartic theorem. For every $\varepsilon>0$, there are integers $M\geq1$ and $a$ such that

$$\limsup_{X\to\infty}\frac M X \#\{n\leq X:n\in S_4,\ n\equiv a\pmod M\}<\varepsilon. \tag{1.1}$$

In particular,

$$\underline d(\mathbb N\setminus S_4)>0. \tag{1.2}$$

The complete assembly of this argument is recorded below. Its new ingredients are the elimination of every principal finite-character quartic coefficient pattern, and a small random omission of primes that suppresses every remaining determinant squareclass. The randomization is only an existence device for a finite modulus.

This does not prove density zero or positive density for $S_4$. No result settling either question for any $r\geq5$ was obtained. Density zero for $S_3$ also remains unresolved here. Thus this is not a complete resolution of all the questions in the problem.

The mathematical argument below has been checked within this investigation; it has not been externally reviewed. One independent internal review checked the elimination and quadratic-character lemmas but not the final analytic note. That review's remaining question was the uniform Gaussian conductor and $L$-function bound, proved in Section 5 below. No specific mathematical gap in the assembled quartic argument was identified.

Two other collaborators read the completed analytic note and checked its Gaussian lattice estimate, conductor argument, trace bookkeeping, omission estimate, and tail bound. Neither reported a substantive gap. This is an internal check, not external confirmation of the new theorem.

2. Canonical decomposition and an exact density-zero reduction

For any $r\geq3$, put

$$\mathcal D_r=\{d\geq1:v_p(d)\in\{0,r+1,\ldots,2r-1\}\text{ for all }p\}.$$

Every $m\in\mathcal P_r$ has a unique expression $m=dx^r$, with $d\in\mathcal D_r$ and $x\geq1$. Indeed, an exponent divisible by $r$ goes entirely into $x^r$; an exponent congruent to $s\in\{1,\ldots,r-1\}$ is uniquely $r+s+ru$, with $u\geq0$. Consequently

$$W_r:=\sum_{d\in\mathcal D_r}d^{-1/r} =\prod_p\left(1+\sum_{s=1}^{r-1}p^{-1-s/r}\right)<\infty, \qquad \#(\mathcal P_r\cap[1,X])\leq W_rX^{1/r}. \tag{2.1}$$

Sums of fewer than $r$ summands therefore account for $O_r(X^{1-1/r})$ integers up to $X$.

For fixed $B$, let $V_{r,B}$ be the union of the value sets of $\sum_{i=1}^r d_ix_i^r$, with $d_i\in\mathcal D_r\cap[1,B]$ and $x_i\geq1$. Counting tuples in a containing box gives

$$0\leq S_r(X)-V_{r,B}(X) \leq rW_r^{r-1}X\sum_{\substack{d\in\mathcal D_r\\d>B}}d^{-1/r} +O_r(X^{1-1/r}). \tag{2.2}$$

For every $0<\eta<1/[r(r+1)]$, the coefficient tail is $O_{r,\eta}(B^{-\eta})$, because $\sum_d d^{-1/r+\eta}<\infty$.

It follows that the following statements are equivalent:

• $S_r$ has density zero;
• every positive diagonal form $a_1x_1^r+\cdots+a_rx_r^r$, with arbitrary fixed positive integer coefficients, has a density-zero value set on positive integer arguments.

For the nontrivial direction involving arbitrary coefficients, multiply the form by $\lambda=(a_1\cdots a_r)^r$. Each $\lambda a_i x_i^r$ is $r$-powerful, so the dilated value set lies in $S_r$. In the other direction use (2.2), first at fixed $B$, then let $B\to\infty$.

In particular, positive upper density of $S_r$ would force positive upper density for at least one fixed critical diagonal form. Allowing infinitely many auxiliary powerful coefficients cannot, by itself, bypass that critical diagonal-form problem. This reduction is not a positive-density proof. A fuller separate version is in positive_density_reduction.md.

For the rest of the record set

$$\mathcal D=\mathcal D_4,\quad F_{\mathbf d}(\mathbf x)=\sum_{i=1}^4d_ix_i^4,\quad P=\prod_{i=1}^4d_i,\quad R=\operatorname{rad}(P),\quad w(\mathbf d)=P^{-1/4}.$$

The squarefree part of $P$ is denoted by $D$; it can equal 1.

Two weighted convergence facts will be used repeatedly. For every fixed $B_0>0$ and $0\leq\eta<1/4$,

$$\sum_{\mathbf d\in\mathcal D^4} w(\mathbf d)B_0^{\omega(R)}R^\eta<\infty. \tag{2.3}$$

The local Euler factor is $1+O_{B_0}(p^{-5/4+\eta})$: the smallest nontrivial coefficient exponent is 5. In particular,

$$\sum_{\substack{\mathbf d\in\mathcal D^4\\P^+(R)>Y}} w(\mathbf d)B_0^{\omega(R)}\ll_{B_0}Y^{-1/8}. \tag{2.4}$$

Here the condition implies $R>Y$, so (2.4) follows from (2.3) with $\eta=1/8$. The tuple $R=1$ never belongs to this tail.

3. Elimination of the principal quartic patterns

Call a tuple $\mathbf d$ exceptional in this record if either

$$P\text{ is a square}, \tag{3.1}$$

or, for some permutation $(i,j,k,l)$,

$$d_i^2d_j^2d_kd_l^3=4t^4\quad\text{for some }t\in\mathbb Q_{>0}. \tag{3.2}$$

This terminology refers to the finite-character expansion below, not to Ghidelli's differently defined exceptional family.

Elimination lemma. If

$$v_2(n)=3,\qquad v_p(n)=1\quad\text{for every odd }p\mid P, \tag{3.3}$$

then an exceptional tuple cannot represent $n$ by $F_{\mathbf d}$, even with arbitrary integer variables.

We first justify the quadratic obstruction. A positive definite rational quaternary diagonal form of square determinant is similar to $\langle1,A,B,AB\rangle$, with $A,B>0$. This is the norm form of $(-A,-B\mid\mathbb Q)$. Its quaternion algebra is division at the real place. If the form were isotropic over every $\mathbb Q_p$, the algebra would split at every finite prime. Hilbert reciprocity forbids exactly one ramified place. This uses precisely Voight, Proposition 14.2.1 and Corollary 14.2.3. The similarity follows directly by dividing a diagonal form $\langle a,b,c,d\rangle$ by $a$ and rescaling its last coordinate, using $abcd\in\mathbb Q^{\times2}$.

At an odd prime, a diagonal ternary form with unit coefficients has a nontrivial zero: the sets $\{aX^2\}$ and $\{-c-bY^2\}$ in $\mathbb F_p$ both have size $(p+1)/2$ and intersect. Set $Z=1$ and lift. Thus a unit-coefficient quaternary form is isotropic there. Also, if $Q(\mathbf z)=n$ and a diagonal term $a_jz_j^2$ is a unit, then $Q$ is isotropic over $\mathbb Q_p$ whenever $p\mid n$ for odd $p$, or $8\mid n$ for $p=2$. Replace $z_j$ by

$$z_j\sqrt{1-n/(a_jz_j^2)}.$$

The required square root exists in $1+p\mathbb Z_p$ for odd $p$, or for arguments in $1+8\mathbb Z_2$ at 2, by Hensel lifting.

Suppose first that (3.1) holds and $F_{\mathbf d}(\mathbf x)=n$. Apply these observations to $Q=\sum d_i z_i^2$, at $z_i=x_i^2$. Each original summand has valuation zero or at least four at every prime. At each odd $p\mid P$, (3.3) forces a unit summand; at 2 the same follows because otherwise $16\mid n$. Hence $Q$ is isotropic at these primes. At every other odd prime its coefficients are units. This contradicts the quaternion obstruction.

For (3.2), write $d_k=cu^2$, $d_l=cv^2$, with $c$ positive squarefree and $u,v$ positive integers. The relation implies

$$d_id_jc^2uv^3=2t^2,\qquad 2d_id_juv=(2t/(cv))^2.$$

Use the identity

$$d_kx_k^4+d_lx_l^4 =c(ux_k^2-vx_l^2)^2+2cuv(x_kx_l)^2. \tag{3.4}$$

The resulting positive quaternary form

$$Q'=\langle d_i,d_j,c,2cuv\rangle$$

has square determinant. At an odd $p\mid P$, a unit among the first two original terms gives a unit term of $Q'$. If neither is a unit, both are divisible by $p^4$. The remaining original terms must both be units: exactly one would make their sum a unit, and neither would make it divisible by $p^4$. Then $2cuv(x_kx_l)^2$ is a unit term of $Q'$. At 2 at least one of the first two terms must be odd. Otherwise the remaining terms would both be odd, and

$$c(u^2x_k^4+v^2x_l^4)\equiv2c\pmod8,$$

contrary to (3.3). All remaining odd primes are unit-coefficient primes for $Q'$. Thus $Q'$ is isotropic everywhere finite, again impossible. This proves the elimination lemma.

Notice that it is a congruence obstruction: if all primes dividing $P$ occur in the prescribed modulus, there are no congruence solutions at all. Indeed any residue-vector solution would give an integer $n=F_{\mathbf d}(\mathbf x)$ satisfying (3.3).

4. A uniform quadratic prime-harmonic estimate

The random omission argument needs a two-sided, conductor-uniform estimate, not just fixed-character convergence:

$$\sup_{z\geq2}\left|\sum_{p\leq z}\frac{\chi(p)}p\right| \ll\log(2q) \tag{4.1}$$

for every nonprincipal real Dirichlet character presented modulo $q$. No Siegel estimate is needed. Here are the details of an elementary weak polynomial lower bound sufficient for (4.1).

For primitive real $\chi$, fixed-character nonvanishing $L(1,\chi)>0$ follows from the fixed-character prime number theorem recorded in Ghidelli, Lemma 5.4: partial summation gives convergence of $\sum_p\chi(p)/p$, hence a finite real logarithm of $L(1,\chi)$. There is no need for uniform constants in this application.

Let $\zeta=e^{2\pi i/q}$. The Gauss sum has absolute value $\sqrt q$. The sine Fourier series gives, for odd $\chi$,

$$|L(1,\chi)|=\pi q^{-3/2}\left|\sum_{a=1}^q a\chi(a)\right| \geq\pi q^{-3/2}. \tag{4.2}$$

The integer sum is nonzero by nonvanishing. For even $\chi$, set

$$U=\prod_{(a,q)=1}(1-\zeta^a)^{\chi(a)}.$$

This is a positive algebraic unit. Indeed the factors $(1-\zeta^a)/(1-\zeta)$ and their inverses are algebraic integers by geometric-sum formulas; the remaining factors cancel since $\sum_a\chi(a)=0$. Pairing $a$ and $-a$ proves positivity. Galois automorphisms send $U$ to $U$ or $U^{-1}$, so $U+U^{-1}$ is an integer. The cosine Fourier series gives $\sqrt q\,|L(1,\chi)|=|\log U|$. Nonvanishing implies $U\ne1$, so $U+U^{-1}\geq3$ and

$$L(1,\chi)\geq q^{-1/2}\log((3+\sqrt5)/2).$$

Both parities give $L(1,\chi)\gg q^{-3/2}$.

Periodicity, zero mean, and partial summation give, uniformly for $1\leq\sigma\leq2$,

$$|L(\sigma,\chi)|\ll\log(2q),\qquad |L'(\sigma,\chi)|\ll\log^2(2q).$$

Near 1, the derivative bound preserves half the preceding lower bound. Outside that short interval use the Euler-factor inequality

$$L(\sigma,\chi)\geq\zeta(2\sigma)/\zeta(\sigma) \gg\sigma-1.$$

Together these give

$$L(\sigma,\chi)\gg q^{-3/2}\log^{-2}(2q) \quad(1<\sigma\leq2). \tag{4.3}$$

For $z\geq3$, set $\sigma=1+1/\log z$. Chebyshev's elementary prime estimates show uniformly in the character that

$$\sum_{p\leq z}\frac{\chi(p)}p=\log L(\sigma,\chi)+O(1). \tag{4.4}$$

The smoothing difference below $z$ is bounded by $(\log z)^{-1}\sum_{p\leq z}(\log p)/p=O(1)$; the absolute tail above $z$ is bounded by a constant times $\int_1^\infty e^{-u}du/u$; prime powers of exponent at least two have bounded total. Equations (4.3)--(4.4) prove (4.1). Removing Euler factors to pass to an imprimitive character costs at most $\sum_{p\mid q}1/p\ll\log(2q)$. Differences of endpoints give the same estimate on every prime interval.

The same smoothing calculation with the principal character gives

$$\sum_{p\leq z}p^{-1}=\log\log z+O(1).$$

Using (4.1) for $\chi_{-4}$, we obtain uniformly for $3\leq Y<Z$

$$\sum_{\substack{Y<p\leq Z\\p\equiv3\ (4)}}p^{-1} =\tfrac12\log(\log Z/\log Y)+O(1). \tag{4.5}$$

A longer proof, including the Gauss-sum identities, is in quadratic_prime_bound.md.

5. Gaussian characters: the required uniformity

The supplied Ghidelli paper provides two inputs used here:

• Proposition 5.2 identifies $\phi_t=(t/\cdot)_4$ as an order-dividing-four Gaussian Hecke character, with some defining ideal supported on $2t$. It identifies the normalized Jacobi character $\mathfrak J_4$, with fixed defining ideal $(16)$, and infinity type $\alpha/|\alpha|$.
• Proposition 4.3 gives the exact quartic zero-class character expansion written in Section 6 below. Kummer theory in Section 6.2 identifies $\phi_t$ as principal precisely when $t\in\mathbb Q(i)^{\times4}$.

The unspecified exponent of a defining ideal in Proposition 5.2 must not silently be assumed uniform. The following argument supplies that uniformity. Since $\mathbb Z[i]$ is a PID, a finite Hecke character can be regarded as a character on the unit group of a residue ring. At an odd prime ideal, principal units modulo a prime power form an odd-primary group, so an order-dividing-four character kills them. At $\pi=1+i$, every unit congruent to 1 modulo $\pi^9$ is a fourth power: Hensel applied at 1 to $T^4-u$ uses $v_\pi(4)=4$ and $v_\pi(u-1)>8$. This also gives a fourth power in every finite quotient. Thus

$$\mathfrak f(\phi_t)\mid (1+i)^9\prod_{\substack{\mathfrak p\mid(t)\\\mathfrak p\nmid2}} \mathfrak p. \tag{5.1}$$

If all rational prime divisors of $t$ divide $2R$, a defining ideal of norm $O(R^2)$ suffices. The same is true for a twist by $\mathfrak J_4^2$, whose infinity type is 2 and whose fixed conductor only changes the absolute constant.

Here is the uniform upper bound for the associated $L$-functions. Let $\psi$ have a defining ideal $\mathfrak m=(\mu)$ of norm $H$, and infinity type $k=0$ or 2. For $k=0$, assume it is nonprincipal. Write

$$\psi((\alpha))=\eta(\alpha)(\alpha/|\alpha|)^k,$$

where $\eta$ is a finite residue character, extended by zero on nonunits. The expression is invariant under Gaussian units. For every residue class modulo $\mathfrak m$, comparison with the square cells of $\mu\mathbb Z[i]$ gives

$$\sum_{\substack{0<|\alpha|\leq T\\\alpha\equiv b\ (\mathfrak m)}} (\alpha/|\alpha|)^k =H^{-1}\int_{|z|\leq T}(z/|z|)^k\,dz +O_k(T/\sqrt H+1). \tag{5.2}$$

Boundary cells give the indicated boundary error. Away from the central cells the angular function has gradient $O_k(1/|z|)$; integrating it over the disk gives the same error. The central cells contribute $O_k(1)$. For $k=2$ the integral vanishes. For $k=0$ its coefficient after summing against $\eta$ vanishes because the finite character is nonprincipal. Dividing the Gaussian integer sum by four yields

$$\sum_{N\mathfrak a\leq x}\psi(\mathfrak a) \ll\sqrt{Hx}+H.$$

Split the Dirichlet series at $H$, and use partial summation for the tail. The initial absolute sum is $O(\log(2H))$, and the tail is $O(1)$, uniformly for $1<\sigma\leq2$. Therefore

$$|L(\sigma,\psi)|\ll\log(2H). \tag{5.3}$$

The prime-ideal version of the smoothing calculation (4.4) gives

$$\sum_{N\mathfrak p\leq z} \frac{\Re\psi(\mathfrak p)}{N\mathfrak p} \leq\log\log(2H)+O(1). \tag{5.4}$$

Its constants are absolute for these two fixed infinity types. Chebyshev's estimates for Gaussian prime ideals follow directly from those for rational primes. Inert prime ideals have norm $p^2$, so discarding them changes such a harmonic sum by $O(1)$.

6. Local counts and the full quartic trace

For an odd prime prescribe the target $p\pmod{p^2}$. Let $\rho_{\mathbf d}(p)$ be the number of solutions modulo $p^2$, divided by $p^6$. If $p\nmid P$, then

$$\rho_{\mathbf d}(p) =1+A_{\mathbf d}(p)(p^{-1}-p^{-2})-p^{-3},\qquad |A_{\mathbf d}(p)|\leq21. \tag{6.1}$$

Indeed the zero-class count modulo $p$ is $p^3+A_{\mathbf d}(p)p(p-1)$. All solutions except the zero vector are nonsingular and have exactly $p^3$ lifts to the specified target modulo $p^2$. The zero vector has none.

For $p\equiv3\pmod4$, fourth powers and squares have the same distribution in $\mathbb F_p$. The quadratic Gauss-sum evaluation therefore gives

$$A_{\mathbf d}(p)=\chi_D(p). \tag{6.2}$$

For $p\equiv1\pmod4$, Ghidelli's exact expansion is

$$\begin{split} A_{\mathbf d}(p)={}&\chi_D(p) +\frac14\sum_{\sigma\in\mathfrak S_4} \chi_4(d_{\sigma(1)}d_{\sigma(2)} d_{\sigma(3)}^3d_{\sigma(4)}^3)\\ &+\frac12\sum_{\sigma\in\mathfrak S_4} \chi_4(-d_{\sigma(1)}^2d_{\sigma(2)}^2 d_{\sigma(3)}d_{\sigma(4)}^3) +2\Re\bigl(\phi_P^{-1}\mathfrak J_4^2\bigr)(\mathfrak p), \end{split} \tag{6.3}$$

where $\mathfrak p$ is one Gaussian prime above $p$ and $\chi_4=\chi_{4,\mathfrak p}$. The finite terms after $\chi_D$ are three conjugate pairs and six conjugate pairs. All multiplicities are positive. Their bounds, together with the last pair, give 21.

A positive rational number is a fourth power in $\mathbb Q(i)$ exactly when it is a rational fourth power. A negative rational number $-A$, with $A>0$, is a fourth power there exactly when $A=4t^4$ for rational $t$. To check this, the imaginary part of $(x+iy)^4$ is $4xy(x^2-y^2)$. Thus a principal character in the second term of (6.3) forces $P$ square, and a principal character in the third term is precisely (3.2). The last character has infinity type 2 and is never principal.

Consequently, for every nonexceptional tuple all the finite characters in this expansion are nonprincipal and $D>1$. Apply (5.4) to the nine finite Gaussian conjugate pairs and the one infinity-type-two pair. Combine the quadratic term $\chi_D(p)$ over all good rational primes, using the elementary upper bound $L(\sigma,\chi_D)\ll\log(2D)$. This last combination is essential: it avoids introducing an inverse quadratic $L$-function in the full baseline. The result is

$$\sum_{\substack{p\leq Z\\p\nmid2P}} \frac{A_{\mathbf d}(p)}p \leq C\log\log(3R)+C\omega(R)+C. \tag{6.4}$$

Removing additional primes dividing $P$ from an Euler sum costs at most $C\sum_{p\mid2P}1/p$, explaining the $\omega(R)$ term.

At a bad coefficient prime, the following bound is crucial:

$$0\leq\rho_{\mathbf d}(p)\leq4\quad(p\mid P). \tag{6.5}$$

Call a coefficient active if it is a unit modulo $p$. If all active variables vanish modulo $p$, the sum is zero modulo $p^2$, because inactive canonical coefficients have valuation at least five. Thus these singular solutions cannot lift to the target $p$. Every other solution is nonsingular and has $p^3$ lifts. Fixing three variables bounds the number of solutions modulo $p$ by $4p^3$, using an active variable; if there is no active variable there are no solutions. This proves (6.5).

At 2 prescribe $8\pmod{16}$, and normalize its solution count by $16^3$. Its local factor is at most 16. These local factors multiply by the Chinese remainder theorem.

7. The random omission construction

Fix a sufficiently small absolute $\delta>0$. For each positive squarefree $E>1$, set

$$\epsilon_E=\frac{\delta}{16E^2},\qquad h(p)=\frac\delta2+\sum_{\substack{E>1\\E\ { m squarefree}}} \epsilon_E\chi_E(p). \tag{7.1}$$

The series converges uniformly and absolutely, $\sum_E\epsilon_E<\delta/4$, and $\sum_E\epsilon_E\log(2E)\ll\delta$. In particular $\delta/4<h(p)<3\delta/4<1$.

Let $Z\to\infty$ and put

$$Y=(\log Z)^{200}.$$

For sufficiently large $Z$, $3<Y<Z$. Include every odd prime $p\leq Y$, and every $p\leq Z$ with $p\equiv1\pmod4$. For each $Y<p\leq Z$, $p\equiv3\pmod4$, independently omit the prime with probability $h(p)$. For the resulting finite set $\mathcal Q$, let

$$M=16\prod_{p\in\mathcal Q}p^2, \qquad a\equiv8\pmod{16},\quad a\equiv p\pmod{p^2} \quad(p\in\mathcal Q). \tag{7.2}$$

The CRT determines $a\pmod M$.

For every fixed positive squarefree $D>1$, (4.1) and (4.5) imply

$$-\sum_{\substack{Y<p\leq Z\\p\equiv3\ (4)}} \frac{h(p)\chi_D(p)}p \leq-\frac{\epsilon_D}{2}\log\frac{\log Z}{\log Y} +C\delta\log(2D)+C\delta. \tag{7.3}$$

To verify this, expand (7.1). The $E=D$ term has $\chi_E(p)\chi_D(p)=1$ off primes dividing $D$, and gives the negative main term. For $E\ne D$, both the product character and its twist by $\chi_{-4}$ are nonprincipal, so their window sums are $O(\log(2DE))$. The same applies to the baseline term $\delta\chi_D/2$. Absolute convergence justifies summing these bounds over $E$. The primes dividing $D$ cost $O(\epsilon_D\log(2D))$. Removing primes dividing $P$ costs at most $C\delta\omega(R)$.

Let $\rho_{\mathbf d}(a,M)$ denote the full normalized solution count modulo $M$, namely the number of residue-vector solutions divided by $M^3$. At an eligible good prime, independence replaces its local factor by

$$h(p)+(1-h(p))\rho_{\mathbf d}(p) =1+(1-h(p))\chi_D(p)/p+O(p^{-2}). \tag{7.4}$$

At other good primes (6.1) applies. At bad primes use 4, and at 2 use 16. The inequality $1+u\leq e^u$, (6.4), and (7.3) give, for each nonexceptional tuple,

$$\mathbb E\rho_{\mathbf d}(a,M) \leq C_0^{\omega(R)+1}(\log(3R))^{C_1}R^{C_2\delta} \left(\frac{\log Z}{\log Y}\right)^{-\epsilon_D/2}. \tag{7.5}$$

All constants are absolute, independent of the tuple, $Y$, and $Z$. Errors $O(p^{-2})$ have an absolutely bounded sum. This argument does not require dividing by a possibly vanishing local factor.

Choose $\delta$ so small that $C_2\delta<1/16$, and absorb the logarithmic power into, for example, $R^{1/16}$. Equation (2.3) now provides a summable majorant for $w(\mathbf d)\mathbb E\rho_{\mathbf d}(a,M)$ over all nonexceptional tuples. For each such fixed tuple, $\epsilon_D>0$ and $\log Z/\log Y\to\infty$, so (7.5) tends to zero. Dominated convergence proves

$$\sum_{\mathbf d\ { m nonexceptional}} w(\mathbf d)\mathbb E\rho_{\mathbf d}(a,M)\longrightarrow0. \tag{7.6}$$

For an exceptional tuple with $P^+(R)\leq Y$, the elimination lemma gives $\rho_{\mathbf d}(a,M)=0$ for every outcome of the random choices. Every remaining exceptional tuple has $P^+(R)>Y$. Uniformly over all choices, (6.1), (6.5), and the bound 21 imply

$$\rho_{\mathbf d}(a,M) \ll 4^{\omega(R)}(\log Z)^{21}.$$

Indeed an included good-prime factor is at most $\exp(21/p+O(p^{-2}))$; an omitted one is 1. Equation (2.4) gives

$$\sum_{\mathbf d\ { m exceptional}} w(\mathbf d)\mathbb E\rho_{\mathbf d}(a,M) \ll (\log Z)^{21}Y^{-1/8} =(\log Z)^{-4}\longrightarrow0. \tag{7.7}$$

Combining (7.6)--(7.7), and using Tonelli for nonnegative terms,

$$\mathbb E\left[\sum_{\mathbf d\in\mathcal D^4} w(\mathbf d)\rho_{\mathbf d}(a,M)\right] \longrightarrow0. \tag{7.8}$$

For each $Z$ there are only finitely many random choices. Thus there is a deterministic outcome making this entire weighted local sum arbitrarily small. This handles the infinite coefficient family, not merely a finite diagonal subsystem.

8. From the weighted local sum to actual density

This step is included explicitly to avoid confusing a narrower equation or a local estimate with the required assertion about $S_4$.

Fix any outcome $(a,M)$ of (7.2). For a fixed tuple $\mathbf d$, lattice-point counting in the compact region $\sum d_i t_i^4\leq X$, $t_i\geq0$, in each residue class gives

$$\#\{\mathbf x\in\mathbb N^4:F_{\mathbf d}(\mathbf x)\leq X, F_{\mathbf d}(\mathbf x)\equiv a\pmod M\} =\frac{VX}{M}w(\mathbf d)\rho_{\mathbf d}(a,M)+o_{M,\mathbf d}(X), \quad V=\Gamma(5/4)^4. \tag{8.1}$$

For completeness, the number in each vector residue class is its region's volume divided by $M^4$, plus $o(X)$, by ordinary Riemann-sum approximation of this Jordan-measurable region after dilation by $X^{1/4}$. The positive-orthant unit $\ell^4$-ball has volume $\Gamma(5/4)^4/\Gamma(2)=V$; scaling coordinate $i$ by $d_i^{-1/4}$ multiplies it by $w(\mathbf d)$. Multiplication by the number $M^3\rho_{\mathbf d}(a,M)$ of allowed vector residues proves (8.1).

For every $X$, the left side of (8.1), divided by $X$, is at most $w(\mathbf d)$, by counting in the containing coordinate box. These weights are summable. Therefore dominated convergence legitimately sums (8.1) over all canonical tuples. The total number of ordered four-term representations in the progression, divided by $X/M$, tends to

$$V\sum_{\mathbf d\in\mathcal D^4} w(\mathbf d)\rho_{\mathbf d}(a,M). \tag{8.2}$$

By (7.8), choose an outcome for which (8.2) is less than $\varepsilon$. Distinct represented integers are no more numerous than ordered representations. Representations with one, two, or three summands account for only $O(X^{3/4})$ integers by (2.1), and $M$ is now fixed. This proves (1.1). Taking, for example, $\varepsilon=1/2$, the progression contains at least $X/(2M)+o(X)$ nonvalues, proving (1.2).

9. Failed approaches and calculations retained

Elementary tuple counting does not settle the problem

For fixed $r$, the ordered tuple count is

$$\Gamma(1+1/r)^rW_r^rX+o(X),$$

by the same finite-coefficient lattice count and summable tails. Dividing by $r!$ gives the unordered leading constant; tuples with equal summands are only $O(X^{1-1/r})$. This does not give an $o(X)$ support estimate. Nor does its constant eventually become less than one: just the Euler factors at 2 and 3 give

$$W_r\geq(1+(r-1)/4)(1+(r-1)/9)\geq r^2/36.$$

Since $\Gamma(1+1/r)\geq e^{-1}$ and $r!\leq r^r$, the unordered constant is at least $(r/(36e))^r$, which tends to infinity.

Simultaneously splitting all quartic coefficients fails

The Fermat quartic has a useful deficit at primes $p\equiv5\pmod8$. But if the canonical core contains $2^5$, making that coefficient a fourth power forces 2 to be a fourth power, hence a quadratic residue, and therefore forces $p\equiv1\pmod8$ among primes $1\pmod4$. Thus the cubic all-coefficients-splitting construction cannot simply be copied in degree four.

A finite quadratic-character mask has a circular tail problem

At primes $3\pmod4$, the trace is $\chi_D(p)$. For finitely many nontrivial squareclasses one can exclude completely split primes in their multiquadratic field; every nontrivial character then has negative mean. However, the saving can be as small as the reciprocal of the size of the squareclass group. The reciprocal-prime mass needed to suppress the core then grows with that group, while the crude uncontrolled tail grows exponentially in the same mass. Choosing the core first does not resolve this circular dependence. The infinite small omission mask in Section 7 replaces this with a uniform integrable majorant.

Eliminating only square determinants is insufficient

The nonsquare determinant tuple $(1,1,1,32)$ has a principal finite character in the third term of (6.3). A baseline proof that assumes all nonsquare-determinant tuples have only nonprincipal finite characters is false. This example is itself excluded by $n\equiv8\pmod{16}$. The general repair is identity (3.4), which eliminates every principal third-term character, not just this example.

Fixed prime-power valuation gaps are not enough

For example, the canonical tuple $(1,1,64,64)$ leads to a positive square-determinant quadratic form, but its quartic values can exhibit every valuation class modulo four at 2: the first pair gives valuations 0 or 1, and the second gives 6 or 7, with fourth-power scaling. Thus one cannot simply assert that quaternion anisotropy produces a missing valuation class. The successful obstruction instead prescribes valuation one at every odd coefficient prime and valuation three at 2, forcing local isotropy of the associated positive square-determinant quadratic form.

Crude inverse $L$-bounds need a small coefficient

A direct selected-prime Euler product can involve inverse quadratic $L$-factors with fixed large exponents. Polynomial lower bounds then produce a coefficient loss too large for the canonical weight. The successful argument first bounds the complete quartic baseline using only positive $L$-powers, and introduces inverse-type estimates solely through an omission probability of size $\delta$. The resulting radical loss is $R^{C\delta}$, which can be made summable. Neither a uniform Siegel bound nor an unproved zero-free hypothesis is being substituted here.

Numerical calculation was diagnostic only

A finite Euler-product computation over primes up to $10^6$ gave approximately $W_4\approx9.5971$. For the weight restricted to square determinant products, the local factor is

$$\tfrac12\bigl[(1+p^{-5/4}+p^{-3/2}+p^{-7/4})^4 +(1-p^{-5/4}+p^{-3/2}-p^{-7/4})^4\bigr].$$

The corresponding truncated product was about 142.515, giving an unordered weighted volume about 4.008. These are truncated numerical values, not certified infinite-product values and not used in the proof. They discouraged the idea that simply retaining all square-determinant tuples would have total representation constant less than one.

10. Remaining uncertainties and continuation points

The density-zero question is untouched by the quartic complement argument. Making one progression very sparse, with its modulus changing with $\varepsilon$, does not imply global density zero. It is equally compatible with positive density of $S_4$. The same distinction applies to the supplied cubic theorem.

No unconditional positive-density construction was found. The exact reduction in Section 2 shows that such a construction would yield positive upper density for some fixed positive diagonal form in the critical number of variables. Conditional six-variable cubic energy bounds from the supplied Wang paper were not treated as unconditional results, and no statement about the narrower powerful equation $a+b=c$ was promoted to a statement about arbitrary target integers in $S_r$.

For $r\geq5$, the quartic mechanism does not automatically extend. The useful zero-class local fluctuations are no longer of divergent reciprocal-prime size, and the four-dimensional positive quadratic norm-form obstruction is special. No replacement for either ingredient was established.

For further checking, the most sensitive parts of the quartic proof are:

1. preserving the full-prime quadratic term when deriving (6.4);
2. eliminating both kinds of principal finite-character pattern, not just square determinants;
3. keeping the conductor estimate (5.1) explicit and uniform;
4. using a two-sided bound in (4.1) for the omission cross terms;
5. retaining the growing support cutoff $Y$, so the exceptional infinite tail is bounded by (7.7), rather than silently discarding it;
6. fixing the final modulus before using the counting limit (8.2).

Each of these has a stated argument above.

Supporting files created during this investigation, none of which is retained here:

• positive_density_reduction.md — the general exact density-zero reduction;
• quartic_elimination.md — the complete quadratic/quaternion elimination;
• quadratic_prime_bound.md — detailed uniform quadratic harmonic estimates;
• quadratic_harmonic_bound.md — an independently written version of that lemma;
• quartic_analytic.md — the uniform Gaussian and randomized local estimates.

The sources actually needed by the quartic proof are Ghidelli's supplied paper (the exact character formula, the Hecke-character correspondence, Kummer theory, and fixed-character nonvanishing) and Voight's supplied chapter (quaternion reciprocity). The canonical decomposition and lattice framework are also developed explicitly above, consistent with the supplied Beyer de Ryke paper. No purported later research note is an input.