Two failures of the one-relation form give two real logarithms of algebraic numbers with algebraic product

Statement. Suppose the relation "$t\neq 0$ with $e^{t}$ algebraic $\Rightarrow$ $t^{2}+\pi^{2}$ transcendental" fails at $t_{1}$ and at $t_{2}$: both $e^{t_{1}},e^{t_{2}}$ are algebraic and both $t_{i}^{2}+\pi^{2}$ are algebraic. Then $t_{1}+t_{2}$ and $t_{1}-t_{2}$ are again real logarithms of algebraic numbers --- $\log\beta_{1}\beta_{2}$ and $\log(\beta_{1}/\beta_{2})$ --- their product

is algebraic, and it is non-zero as soon as $t_{1}^{2}\neq t_{2}^{2}$.

Read contrapositively: if the product of two $\mathbb{Q}$-linearly independent real logarithms of algebraic numbers is always transcendental, the relation can fail at only one value of $|t|$. Nothing here proves the relation; it bounds how badly it could fail.

Source and attribution. All the mathematics of this section is Carlo Perassi's, in his manuscript C. Perassi, Rigidity of logarithms with algebraic modulus — Around a conjecture of Diaz, unpublished manuscript, 15 August 2026, §Polar coordinates and the discreteness of the period. No novelty is claimed. The configuration produced here --- two $\mathbb{Q}$-linearly independent $\lambda_{1},\lambda_{2}\in\mathcal{L}$ with $\lambda_{1}\lambda_{2}\in\bar{\mathbb{Q}}^{\times}$ --- is exactly condition eq:exotic of Theorem Period-plane classification, and the manuscript's Remark What the model sees, and the exact gap (rem:period-plane-scope) calls its solvability "the one undecided point", notes that the strong four exponentials conjecture excludes it, and that Roy's strong six exponentials does not reach it. The manuscript raises eq:exotic with purely imaginary logarithms, which is why Diaz's assertion (C6) of 1997 --- whose hypothesis is $|\alpha|\neq 1$ --- does not apply there (Remark Adjacent results, rem:exotic-adjacent). Here the logarithms are real, so $|\alpha|\neq 1$ does hold. That the failure set of the one-relation form feeds the same condition with real logarithms is possibly new and is small; it is possibly folklore, and it was not found stated in the sources consulted.

Proof. $e^{t_{1}+t_{2}}=e^{t_{1}}e^{t_{2}}$ and $e^{t_{1}-t_{2}}=e^{t_{1}}/e^{t_{2}}$, and the algebraic numbers are closed under product and quotient. The product identity is a subtraction, and $t_{1}^{2}-t_{2}^{2}=0$ is $t_{1}^{2}=t_{2}^{2}$.