A same-conductor squaring identity with a nontriviality certificate

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Standing and exact hypotheses

Suppose $w$ and the distinct roots in $B$ all have odd order, $w\notin B$, and

$$2w=\sum_{b\in B}\epsilon_b b,\qquad \epsilon_b\in\{\pm1\}. \tag{1}$$

Unlike the conductor-disjoint determinant construction, this lemma works inside one conductor. It gives a nonzero integer relation on pair products and squares when $|B|>4$. It does not automatically give a ternary relation: multiplicative collisions can increase coefficients. Nor does it control the actual colors of the resulting roots.

These are precisely the surviving obligations. No general-height extension, ternary-subrelation theorem, or global coloring conclusion is asserted.

1. The squaring identity

Choose one odd $N$ containing the conductors in (1). Squaring is a cyclotomic Galois automorphism of $\mathbb Q(\zeta_N)$, so

$$\sum_b\epsilon_b b^2=2w^2.$$

Squaring (1) itself gives $(\sum_b\epsilon_b b)^2=4w^2$. Subtract the preceding equation and divide by two to obtain

$$\boxed{ w^2=\sum_{\epsilon_b=-1}b^2+ \sum_{\{b,c\}\subset B,\ b\ne c}\epsilon_b\epsilon_c\,bc. } \tag{2}$$

The second sum is over unordered pairs, counted once. Before identifying equal root products, every listed term has coefficient $+1$ or $-1$. Squares occur only for negative leaves.

To distinguish that indexed expression from the actual relation, work in the group ring $\mathbb Z[\mu_N]$, writing $[g]$ for its formal basis element at $g$. Put

$$U=\sum_b\epsilon_b[b],\qquad \psi_2(U)=\sum_b\epsilon_b[b^2],\qquad \lambda_2(U)=\frac{U^2-\psi_2(U)}2.$$

The last expression has integer coefficients by its explicit expansion in (2). Evaluation $[g]\mapsto g$ sends

$$R=\lambda_2(U)-[w^2] \tag{3}$$

to zero. Colliding pair products, squares, and $w^2$ must be collected in this group ring before discussing their coefficient height.

2. Nontriviality after all multiplicative collisions

Proposition. If $|B|>4$, then $R\ne0$ in $\mathbb Z[\mu_N]$.

Multiply every root in (1) by $w^{-1}$, so $w=1$. This multiplies (3) by the invertible group-ring element $[w^{-2}]$; it does not change formal nontriviality or the number of distinct leaves. Suppose for contradiction that $\lambda_2(U)=[1]$. Then

$$\psi_2(U)=U^2-2[1]. \tag{4}$$

For each character $\chi$ of $\mu_N$, write $z_\chi=\chi(U)$. Equation (4) gives

$$z_{\chi^2}=z_\chi^2-2.$$

Since $N$ is odd, $\chi\mapsto\chi^2$ permutes the finite character group. Every $z_\chi$ is therefore a periodic point of $F(z)=z^2-2$.

Every complex periodic point of $F$ lies in $[-2,2]$. For a direct proof, choose $t\ne0$ with $z=t+t^{-1}$. Induction gives

$$F^n(z)=t^{2^n}+t^{-2^n}.$$

If $F^n(z)=z$, the two roots of $X^2-zX+1$ show that $t^{2^n}$ is either $t$ or $t^{-1}$. Hence $t$ is a root of unity, so $z=2\operatorname{Re}t\in[-2,2]$.

Writing $U=\sum_g a_g[g]$, character orthogonality now gives

$$|B|=\sum_{g\in\mu_N}|a_g|^2 =\frac1N\sum_\chi|z_\chi|^2\le4,$$

contrary to the hypothesis. This proves the proposition. In particular it applies to the nineteen-leaf coefficient-two circuits already constructed in the prime grids, without rerunning their certificates.

3. What would make this an actual-color flattening step

If the collected nonzero coefficients of $R$ all lie in $\{\pm1\}$ and their root support has one actual color, (3) is the required monochromatic ternary relation. A stronger sufficient collision condition is that the indexed roots in (2), together with $w^2$, are all distinct; then the coefficients are automatically ternary.

Neither condition has been established for the root supports that a finite coloring can supply. Refining a color by the pair $(f(x),f(x^2))$ controls square colors, but not the colors $f(bc)$ of pair products. Nontriviality from Section 2 prevents total formal cancellation; it does not prevent surviving coefficients of magnitude greater than one.

Thus (2) is a proved same-conductor algebraic alternative, with an exact nontriviality certificate, not a proof of height descent or unbounded coloring. The identity and character argument received independent internal checks. No computation or additional literature result was used.