An exact obstruction to the restricted quartic residue objective

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Status and scope

Section 8 of the independent record claimed a finite obstruction to one quartic construction. Its referenced script was not supplied. The calculation is reconstructed in the preserved exact certificate, and its formulas and rounding are checked. It proves that the specific objective below exceeds $27/25=1.08$ for every choice of nonnegative budgets. This is not an obstruction to the quartic complement argument.

The restricted objective

Fix target $15\pmod{16}$. A canonical coefficient $d=2^eu$ has type $s=(E,v)$, where $e\in\{0,5,6,7\}$, $E=e\pmod4$, and $v=u\pmod{16}$ is odd. There are $32$ types. Write

$$Z_v=\sum_{\substack{u\text{ odd canonical}\\u\equiv v\ (16)}}u^{-1/4}, \qquad a_E=2^{-e/4},\qquad V=\Gamma(5/4)^4.$$

The $32$ character prescriptions are indexed by a character $\psi$ of $(\mathbb Z/16\mathbb Z)^\times$ and $t\in\{0,1,2,3\}$, with $\chi(\ell)=\psi(\ell\bmod16)$ at odd coefficient primes and $\chi(2)=i^t$. The associated sign is

$$\epsilon=\chi(-1)=\chi(2)^2=(-1)^t,$$

not necessarily $\psi(15)$: the prescription concerns positive odd primes. For a multiset of four types, put $z_j=i^{tE_j}\psi(v_j)$ and $D=\prod_jz_j$. The finite part of its zero-target trace is

$$B=D^2\left[1+\epsilon\left(\left|\sum_jz_j\right|^2-4\right)\right] +D\sum_{j<k}z_j^2z_k^2.$$

The remaining Jacobi term has absolute value at most $2$, by the checked Gauss-sum expansion.

Among the $k$ odd coefficients, let $P$ be the fraction of their $2^k$ subsets whose residues sum to $15\pmod{16}$. All even coefficients vanish modulo $16$, while an odd-coefficient term contributes either $0$ or its coefficient residue with equal probability. Thus $\rho_{16}=16P$. If $m_s$ is the multiplicity of type $s$, its relative unordered representation weight is

$$w=16VP\frac{\prod_s(a_EZ_v)^{m_s}}{\prod_s m_s!}.$$

Indeed, type permutations give $4!/\prod_s m_s!$, followed by the usual division by $4!$; representations with repeated actual summands are $o(X)$.

The proposed safe upper-bound objective is

$$\mathcal F(\boldsymbol\lambda) =\sum_{\text{type multisets}}w \exp\left(\sum_g\lambda_g(B_g+2)\right), \qquad\lambda_g\ge0.$$

The finite certificate

The certificate enumerates exactly $\binom{35}{4}=52360$ multisets and all $32$ prescriptions. Exactly $364$ multisets have $P>0$ and $B_g+2\ge0$ for every prescription. Their contributions cannot decrease as any budget grows.

Their weights are bounded below using only odd canonical coefficients supported on the $53$ odd primes at most $251$. The cube $[0,1/\sqrt2]^4$ lies in the positive quartic unit ball, so $V\ge1/4$. All local weights $p^{-e/4}$ and all residue convolutions are rounded down to multiples of $10^{-12}$, using integer fourth-root comparisons. Nonnegative coefficients preserve the direction of every bound. The final sum uses exact rational arithmetic and equals

$$\frac{67826836205251839956626671887649519872177425279} {62500000000000000000000000000000000000000000000} >\frac{27}{25}.$$

This is a newly reconstructed lower bound; no reliance is placed on the unavailable original script or on floating-point optimization.

Hence $\mathcal F(\boldsymbol\lambda)>27/25$ for all nonnegative budgets. The result does not cover other residue targets, other character prescriptions, sharper use of Jacobi phases, additional small-prime restrictions, or distinct-value estimates. In particular, the successful proof first eliminates exceptional tuples using $8\pmod{16}$ and valuation-one targets, then uses a different multiquadratic Frobenius distribution. See the comparison of the two methods.