Quartic finite families and the remaining tail problem

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Status

The finite-family theorem below is proved. By itself it applies only to fixed canonical truncations. The obstruction at the end of this note was subsequently overcome by the character distribution and uniform Euler majorant in the full quartic proof, which establishes positive lower density of the complement of $S_4$.

Theorem

Let $F_1,\ldots,F_J$ be a finite family of forms

$$F_j(\mathbf x)=\sum_{i=1}^4d_{ji}x_i^4,$$

whose coefficients are positive 4-powerful integers. For every $\varepsilon>0$, there is a residue class $a\pmod M$, with $a\equiv8\pmod{16}$, such that

$$\limsup_{X\to\infty}\frac M X \sum_{j=1}^J\#\{\mathbf x\in\mathbb N^4:F_j(\mathbf x)\le X, F_j(\mathbf x)\equiv a\pmod M\} <\varepsilon. \tag{1}$$

In particular, the complement of the union of their positive value sets has positive lower density.

Elimination of square products of coefficients

Let $P$ be the finite set of odd primes dividing some coefficient in the family. Put

$$M_0=16\prod_{p\in P}p^2,$$

and choose $a_0$ by

$$a_0\equiv8\pmod{16},\qquad a_0\equiv p\pmod{p^2}\quad(p\in P). \tag{2}$$

We claim that (2) excludes every form with $\Delta=\prod_i d_i$ a rational square.

Associate the positive definite quadratic form $Q(\mathbf y)=\sum_i d_i y_i^2$. Since its determinant is a square, it is similar over $\mathbb Q$ to

$$\langle1,u,v,uv\rangle,\qquad u,v>0,$$

the norm form of the quaternion algebra $(-u,-v)$. This algebra is ramified at the real place. Hilbert reciprocity implies ramification at at least one finite place. Here we use the standard reciprocity theorem for Hilbert symbols, and the equivalence between splitting of a quaternion algebra and isotropy of its norm form.

If $F(\mathbf x)\equiv8\pmod{16}$, at least one $d_i x_i^4$ is odd: otherwise every term is divisible by 16, because each even coefficient is 4-powerful. Set $y_i=x_i^2$. Then $v_2(Q(\mathbf y))=3$, while the derivative in a coordinate with $d_i,y_i$ odd has valuation 1. The one-variable Hensel criterion $3>2\cdot1$ proves that $Q$ is isotropic over $\mathbb Q_2$.

Consequently $Q$ is anisotropic at some odd prime $p$. This prime divides a coefficient: a four-variable nondegenerate quadratic form with unit coefficients has a nonsingular zero modulo every odd prime and hence is isotropic locally.

At this anisotropic prime, at most two coefficients are units. Three unit coefficients would already give a nonsingular zero of the corresponding ternary quadratic form modulo $p$. If exactly two are units, their binary reduction must be anisotropic. Therefore $F(\mathbf x)\equiv0\pmod p$ forces every variable belonging to a unit coefficient to be divisible by $p$. Their fourth powers are divisible by $p^4$, and every nonunit coefficient is itself divisible by $p^4$. Hence $p^4\mid F(\mathbf x)$, contradicting $F(\mathbf x)\equiv p\pmod{p^2}$. The cases of at most one unit coefficient give the same contradiction.

This proves the claim.

Simultaneous savings for nonsquare products

Discard the excluded forms. Let

$$K=\mathbb Q(\sqrt{\Delta_j}:1\le j\le J),\qquad G=\operatorname{Gal}(K/\mathbb Q).$$

Every remaining $\Delta_j$ is nonsquare and defines a nontrivial quadratic character $\chi_j$ of $G$. The field $K$ is totally real, so $K\cap\mathbb Q(i)=\mathbb Q$.

For each $g\in G\setminus\{1\}$, Chebotarev supplies a positive-density set of primes $q\equiv3\pmod4$ with Frobenius $g$ in $K$, avoiding the finite coefficient-prime set. Its reciprocal sum diverges. Given $T>0$, choose finitely many such primes with

$$T\le\sum_{q\in\mathcal Q_g}\frac1q\le T+\frac1{|G|}. \tag{3}$$

All selected primes can be required to exceed any fixed threshold.

At $q\equiv3\pmod4$, the fourth-power map and the square map on $\mathbb F_q$ have identical fiber cardinalities over each target. The elementary four-variable quadratic Gauss-sum formula therefore gives

$$\rho_q(0;F_j):=q^{-3}\#\{\mathbf x\bmod q:F_j(\mathbf x)=0\} =1+\chi_j(g)\frac{q-1}{q^2}. \tag{4}$$

Using $\log(1+t)\le t$, equations (3)–(4), and $\sum_q q^{-2}<\infty$, we obtain

$$\log\prod_{g\ne1}\prod_{q\in\mathcal Q_g}\rho_q(0;F_j) \le T\sum_{g\ne1}\chi_j(g)+O(1) =-T+O(1). \tag{5}$$

The error is bounded independently of $T$ and $j$ for this fixed family. The final equality holds because $\chi_j$ is nontrivial.

Take

$$M=M_0\prod_g\prod_{q\in\mathcal Q_g}q,$$

and choose $a\equiv a_0\pmod{M_0}$, $a\equiv0\pmod q$ at every selected prime. The local factors at $M_0$ are fixed. By (5), every surviving form's normalized congruence count tends to zero as $T\to\infty$.

Passage from congruence counts to representations

For each fixed form and modulus, positive lattice-point counting gives

$$\#\{\mathbf x\in\mathbb N^4:F_j(\mathbf x)\le X, F_j(\mathbf x)\equiv a\pmod M\} =\frac{\Gamma(5/4)^4X}{M\Delta_j^{1/4}}\rho_M(a;F_j) +O_{M,F_j}(X^{3/4}),$$

where $\rho_M=M^{-3}\#\{\mathbf x\bmod M:F_j(\mathbf x)\equiv a\}$ and CRT makes $\rho_M$ multiplicative. Sum this identity over the finite family, and choose $T$ sufficiently large. This proves (1).

Why this is not yet a theorem about $S_4$

Every fixed canonical truncation satisfies (1). However, the number of quadratic Frobenius classes used is $|G|-1$, and the total reciprocal-prime sum in (3) is about $(|G|-1)T$. A core tuple outside the controlled family can acquire a fallback factor of order $\exp((|G|-1)T)$. Enlarging the core to reduce its weighted tail may enlarge $G$ drastically.

Thus the supplied cubic two-tail proof cannot simply be copied: its fallback amplification is independent of the size of the controlled core, whereas that uniformity has not been obtained here. Nor does positive complement density for every finite truncation imply positive complement density or infinitude for their infinite union.

This obstruction is resolved in the full quartic proof. The distribution $f(g)=1-\sum_m c_m\chi_m(g)$, with fixed positive summable $c_m$, gives class-dependent savings with a total reciprocal-prime budget independent of the Galois rank. The remaining majorant requires eliminating every principal quartic trace character, a larger class than square products alone.

Comparison with the independent quartic obstructions

Sections 5–8 of the independent record describe genuine failures of more restricted approaches. Its conclusion that no quartic infinitude theorem was obtained describes that separate investigation, not the current research state. The checks below do not change the quartic complement argument.

Positive zero-target factors do occur

For the canonical tuple $(1,1,64,64)$, every odd-prime zero-target factor is strictly greater than one. At $p\equiv3\pmod4$ its coefficient product is square, giving $1+(p-1)/p^2$. At $p\equiv1\pmod4$, put $\epsilon=\chi(-1)$ for a quartic character $\chi$, and $J=J(\chi,\chi)$. The checked Gauss-sum expansion gives

$$\rho_p(0)=1+\frac{p-1}{p^2}\left(B+2\Re(J^2/p)\right), \qquad |J^2/p|=1.$$

Since $\chi(64)=(2/p)$, the finite trace $B$ is $19$ when $p\equiv1\pmod8$, and $11$ when $p\equiv5\pmod8$. Thus the traces lie in $[17,21]$ and $[9,13]$, respectively. The pattern calculation agrees with the Euler-majorant expansion.

This example is nevertheless excluded immediately by the successful target $8\pmod{16}$: its values are only $0,1,2\pmod{16}$. Replacing the last coefficients by $4,4$ preserves odd-prime local counts but requires even new root variables. It is not an unrestricted integral or $2$-adic equivalence. The analogous normalization of $(1,32,64,128)$ also does not justify a universality claim.

More generally, the independent paired family consists of tuples equivalent, under permutation, common positive rational scaling and individual rational fourth-power scaling, to $(1,4,a,4a)$. Its good odd-prime zero-target factors are all greater than one. For split primes, $\chi(4)=\epsilon$ gives pattern $(1,\epsilon,z,\epsilon z)$: the finite traces are $19,7,3$ when $\epsilon=1$, and $11,7$ when $\epsilon=-1$. The remaining Jacobi term has absolute value at most $2$. At inert primes the coefficient product is square, giving the positive quadratic trace.

Every paired tuple has square product, so it belongs to the class eliminated by the reciprocity argument above. That argument uses $8\pmod{16}$ and valuation-one targets at the coefficient primes, not only homogeneous conditions at good primes. No reduction of all representations to the paired family is needed.

A checked global bound for the paired subfamily

The independent global bound is correct, although unnecessary for the quartic theorem. For canonical coefficients, paired partners have the same odd part and dyadic exponents $\{0,6\}$ or $\{5,7\}$. Let $\mathcal P$ be the values represented by tuples admitting a partition into two such pairs. Set

$$L=1/\sqrt2+1/4,\quad W=\prod_{p\text{ odd}}(1+p^{-5/2}+p^{-3}+p^{-7/2}),\quad Z=\prod_{p\text{ odd}}(1+p^{-5}+p^{-6}+p^{-7}).$$

A fixed ordered pair has total coefficient weight $LW$. Two different pair partitions force all four odd parts to agree; the dyadic exponents alternate between an allowed pair, giving weight

$$\left(2\cdot2^{-3}+2\cdot2^{-6}\right)Z=\frac9{32}Z.$$

All three partitions cannot hold, since three dyadic exponents cannot have pairwise differences $2\pmod4$. Inclusion–exclusion therefore gives exact ordered coefficient weight

$$3L^2W^2-\frac{27}{32}Z.$$

Lattice counting and the summable coefficient weights give the corresponding ordered tuple asymptotic with factor $VX$, $V=\Gamma(5/4)^4$. Repeated actual summands contribute $O(X^{3/4})$; all other permutation orbits have size $24$. Consequently

$$\overline d(\mathcal P) \le\frac V{24}\left(3L^2W^2-\frac{27}{32}Z\right)<\frac14.$$

For the strict inequality, $V\le1$, $L<1$, and, with $f(t)=t^{-5/2}+t^{-3}+t^{-7/2}$,

$$\log W\le f(3)+\tfrac12\int_3^\infty f(t)\,dt =\frac7{108}+\frac{38}{135\sqrt3}<\frac14.$$

Thus $W^2<\sqrt e<2$. This global upper density does not remain a relative density bound after restricting to an arbitrarily sparse progression. The separate reported bound $<1/200$ inside $15\pmod{16}$ depends on a finite certificate not supplied with the independent record; its numerical step has not been adopted here.

Why the independent tail obstruction is bypassed

For target zero modulo $p$, a tuple with all four coefficients divisible by $p$ has local factor $p$. A majorant ignoring this common-divisor factor is indeed invalid. The independent repair involving $\operatorname{rad}(\gcd(d_1,d_2,d_3,d_4))$ does give a summable majorant uniformly for a fixed reciprocal-prime budget, but not uniformly as that budget grows.

The successful proof instead imposes $F\equiv p\pmod{p^2}$. If all coefficients are divisible by $p$, there are no solutions. Otherwise every solution has a unit-coefficient unit variable, since canonical nonunit coefficients and fourth powers of nonunit variables are divisible by $p^4$. Nonsingular lifting and the one-variable root bound give $\rho_{p^2}(p;\mathbf d)\le4$ at every coefficient prime. This explains the smaller summable weight $4^{\omega(d_1d_2d_3d_4)}$ in the current proof.

For the regular tuples, the uniform Euler majorant is independent of the small-prime cutoff. The later Frobenius distribution $f(g)=1-\sum_m c_m\chi_m(g)$ has total reciprocal-prime budget at most $T+1$, independent of the finite coefficient set. One first chooses $T$ using dominated convergence, then the coefficient-prime cutoff $B$, and finally the finite core $E$. Thus the fallback factor $e^{T+1}$ is fixed before the tails are cut off. The independent record did not establish these ingredients. The proof also eliminates mixed principal-character cases, not just square products; its treatment of the remaining Jacobi term requires an upper bound for a Hecke $L$-function, not angular prime equidistribution.

The independent record's more specific $32$-prescription optimization failure has also been reconstructed as an exact finite obstruction. It applies only to that target-$15$ safe objective, not to the construction used in the current proof. The broader finite-character classification and the exploratory phase optimizations in the independent record are not needed here and have not been incorporated as established results.

Comparison with the later random omission record

The second independent record does obtain the full quartic theorem. Its checked random omission construction is an alternative to the established bounded-budget distribution, not a stronger conclusion. Its warning about a finite quadratic-character mask concerns the equal-budget, nonidentity-Frobenius construction above. It is not a failure of every finite-character method: the distribution $f(g)=1-\sum_m c_m\chi_m(g)$ already resolves that circularity.

Three diagnostic failures from its Section 9 are useful and check directly.

Square-product elimination alone does not justify the Euler majorant. The canonical tuple $(1,1,1,32)$ has nonsquare product, but its mixed monomial can be $32^2=4\cdot4^4$. It therefore has a principal mixed quartic character. Its values are only $0,1,2,3\pmod{16}$, so the successful target $8\pmod{16}$ excludes it. The general mixed-pattern elimination, not merely square determinants, is necessary in the Euler-product proof.

Quaternion anisotropy does not imply a missing valuation class modulo four. For $F=x_1^4+x_2^4+64x_3^4+64x_4^4$, the positive inputs

$$(1,2,1,1),\quad(1,1,1,1),\quad(4,4,1,2),\quad(4,4,1,1)$$

give $145,130,1600,640$, with $2$-adic valuations $0,1,6,7$. Every class modulo four occurs. This does not contradict exclusion of the exact valuation three; indeed the values are only $0,1,2\pmod{16}$. The successful argument forces local isotropy at the specified exact valuations rather than asserting a general valuation-class gap.

Making all coefficients fourth powers can remove the useful primes. At primes $p\equiv5\pmod8$, the Fermat quartic's finite trace is $1+6-12=-5$, so its complete trace is at most $-3$. But a coefficient $2^5$ being a fourth power forces $2$ itself to be a fourth power. Among split primes this implies $(2/p)=1$, hence $p\equiv1\pmod8$. Thus the cubic all-coefficients-splitting strategy cannot simply retain this Fermat deficit in degree four. Neither quartic completion uses that strategy.

The independent numerical Euler products are explicitly uncertified diagnostics and are not adopted as proof inputs. The existing target-$15$ finite certificate concerns a different restricted objective and obstructs neither proof.

Odd-valuation support and a sharper exceptional tail

Appendix B of third independent record correctly strengthens square-product elimination: only odd primes having an odd valuation in some coefficient need be prescribed. Square factors at arbitrarily large primes do not require new congruences. The same refinement also holds for the mixed principal-character cases, although the independent Appendix C states only full-support elimination.

To check this, let $P$ contain all odd primes having an odd coefficient valuation, and use (2). At primes in $P$ and at $2$, the established local lifting arguments are unchanged. At an odd prime outside $P$, all coefficient valuations of $Q=\langle d_1,d_2,d_3,d_4\rangle$ are even; rescaling its coordinates removes those valuations and gives a unit form, which is isotropic.

For a mixed case write

$$d_k=cu^2,\quad d_l=cv^2,\quad d_i^2d_j^2d_kd_l^3\in4\mathbb Q^{\times4}.$$

Here $c$ is positive squarefree and $u,v$ are positive integers. The established identity gives the square-determinant form $Q'=\langle d_i,d_j,c,2cuv\rangle$ and $d_i d_j uv=2s^2$ for some rational $s$. At an odd prime outside $P$, $c$ is a unit and the last identity makes $v_p(uv)$ even. Thus every coefficient of $Q'$ has even valuation too, so coordinate rescaling gives a unit form. All finite places are isotropic, contradicting the same quaternion reciprocity theorem. This proves the refinement for the entire exceptional family of the full quartic proof, not just square products.

There is also a sharper tail for canonical tuples $\mathbf d\in\mathcal C_4^4$. For fixed $\kappa\ge1$, put

$$w_\kappa(\mathbf d)= (d_1d_2d_3d_4)^{-1/4}\kappa^{\sum_i\omega(d_i)}.$$

Then

$$\sum_{\substack{\mathbf d\text{ exceptional}\\ \exists p>Y,\ i:\ v_p(d_i)\text{ odd}}} w_\kappa(\mathbf d)\ll_\kappa Y^{-3/2}\qquad(Y\ge3). \tag{6}$$

For square-product tuples an odd local exponent occurs at least twice. For a mixed tuple, its exponents $e_h\in\{0,5,6,7\}$ at an odd prime satisfy

$$2e_i+2e_j+e_k+3e_l\equiv0\pmod4.$$

A vector having just one nonzero exponent, that exponent odd, cannot satisfy this congruence. Thus every exceptional tuple in the event at $p$ has at least two nonzero local exponents, of total at least ten. Its local weight is $O_\kappa(p^{-5/2})$ even if the exceptional condition is dropped at all other primes. The unrestricted Euler product of the other local weights converges. A union bound and $\sum_{p>Y}p^{-5/2}\ll Y^{-3/2}$ prove (6).

Prescribe all odd primes at most $Y$. Survivors in the exceptional family must belong to (6), whereas their local factors are at most $C4^{\omega(d_1d_2d_3d_4)}(\log(2Y))^{21}$ by the checked trace bound. Their total weighted local mass is consequently $O(Y^{-3/2}(\log(2Y))^{21})=o(1)$. This controls the entire exceptional stratum without a Hecke bound; regular tuples still need the separate suppression argument.

The independent Appendix B uses the weaker elementary logarithmic exponent $81e$, while its summary writes $81$. Either exponent suffices for its conclusion; the existing trace and prime-harmonic estimates give the exponent $21$ used here.

A negative one-prime average cannot be iterated

Section 6 of the third record averages over four independent coefficient choices, each with dyadic weights $1,2^{-5/4},2^{-6/4},2^{-7/4}$, at a prime $q\equiv5\pmod8$, with $\chi_4(2)=i$ and all odd coefficient parts fourth residues. Its displayed mean formula is the direct average of the nineteen finite-character terms. It is an average over tuples, not independence of the local factors as the prime varies.

The fixed positive-weight tuple $(1,1,64,64)$ already proves the failure of such iteration: at every prime with this prescription it has finite trace $11$, hence total trace at least nine. Its contribution grows when the reciprocal-prime budget diverges, despite a negative one-prime mean over all tuples. This tuple is excluded by $8\pmod{16}$ (and also by $15\pmod{16}$). Conditioning on either dyadic target changes the weights, so this example obstructs only the unconditioned averaging argument, not the successful proof or the separate target-$15$ certificate.

The finite-core circularity described in the same independent section is likewise the old equal-budget obstruction, already resolved by the bounded-budget distribution. The newly checked universal selector is another resolution, not evidence that every finite-character construction fails.