Result: An analog to the Schur-Siegel-Smyth trace problem

For a totally positive algebraic integer \(\alpha\), let \(\deg (\alpha^{2})=\deg (\alpha)\) and \(\operatorname{trace}(\alpha^{2})\) be the number and the sum of the conjugates of \(\alpha^{2}\), respectively. Since \[ \lim_{p\rightarrow \infty }\frac{\operatorname{trace}((2+2\cos (2\pi /p))^{2})}{\deg (2+2\cos (2\pi /p))}=6, \] where \(p\) is a prime number, the smallest limit point, say \(l\), of the quantities \(\operatorname{trace}(\alpha^{2})/\deg (\alpha)\), when \(\alpha\) runs through the set of totally positive algebraic integers, is at most \(6\). According to [\textit{Y. Liang} and \textit{Q. Wu}, J. Aust. Math. Soc. 90, No. 3, 341--354 (2011; Zbl 1290.11143)], \[ \frac{\operatorname{trace}(\alpha^{2})}{\deg (\alpha)}\geq 5.31935, \] except when \(\alpha\) belongs to a finite set of explicitly given numbers, and the author of the paper under review uses some explicit auxiliary functions (like in previous works on related problems) to show that this result continues to hold when the constant \(5.31935\) is replaced by \( 5.321767\). By following an approach of J. P. Serre about the optimality of the auxiliary functions method on the trace problem [\textit{J. Aguirre} and \textit{J. C. Peral}, Lond. Math. Soc. Lect. Note Ser. 352, 1--19 (2008; Zbl 1266.11113)], she also proves that the best lower bound for \(l\), which may be obtained by this method, is \(5.895237\ldots\), provided \(l\in (5.895237\ldots,6]\).