At a candidate u, with Roy's strong six exponentials theorem as hypothesis: if a₁ ≠ a₂ are algebraic with aᵢāᵢ = |u|⁴, then u/(u² − a₁) + u/(u² − a₂) ∉ ℒ̃

Here $\mathcal{L}$ is the set of logarithms of algebraic numbers, and $\widetilde{\mathcal{L}}$ is the $\overline{\mathbb{Q}}$-vector space spanned by $1$ and $\mathcal{L}$. Roy's strong six exponentials theorem is carried as the hypothesis hSSE: if $x_1, x_2$ are $\overline{\mathbb{Q}}$-linearly independent and so are $y_1, y_2, y_3$, then one of the six products $x_iy_j$ is not in $\widetilde{\mathcal{L}}$.

Let $u$ be a candidate for Diaz's conjecture (DiazModulus.IsCandidate), $\rho = u\bar u$, and $a_1 \neq a_2$ non-zero algebraic numbers with $a_1\bar a_1 = a_2\bar a_2 = \rho^2$. Then $u/(u^2 - a_1) + u/(u^2 - a_2) \notin \widetilde{\mathcal{L}}$. Here a hypothesis $z \in \widetilde{\mathcal{L}}$ on the sum brings $\bar z$, and the two together give both terms. It concerns candidates, so it is vacuous if Diaz's conjecture holds.

Proof. Since $\rho^2/\bar a_i = a_i$, conjugation acts on each $z_i$ by a scalar: $\bar z_i = k_iz_i$ with $k_i = -\rho/\bar a_i$, and $k_1 \neq k_2$. If $z = z_1 + z_2 \in \widetilde{\mathcal{L}}$, then $\bar z = k_1z_1 + k_2z_2 \in \widetilde{\mathcal{L}}$ (DiazModulus.logAlgTilde_conj_stable), so $z_1 = (\bar z - k_2z)/(k_1 - k_2)$ and $z_2$ are in $\widetilde{\mathcal{L}}$, against DiazModulus.candidate_two_pole_not_both_mem_logAlgTilde.

Novelty. Not asserted: one line from DiazModulus.candidate_two_pole_not_both_mem_logAlgTilde, itself a substitution in Diaz (2007, Th. 7(2)).