Squarefree parts of discriminants of trinomials (Q2510687)

Let \(\Delta(n)\) be the discriminant of the trinomial \(X^n-X-1\), so that \[ \Delta(n)=n^n+(-1)^n(n-1)^{n-1}. \] \textit{D. W. Boyd} et al. [LMS J. Comput. Math. 18, 148--169 (2015; Zbl 1329.11103)] conjectured that \(\Delta(n)\) is square-free for a positive proportion of \(n\), but this seems a long way off. It is shown here that if \(T(N,s)\) is the number of \(n\leq N\) for which the square-free part of \(\Delta(n)\) is \(s\), then for any fixed \(\varepsilon>0\) one has \[ T(N,s)\ll_{\varepsilon} N^{3/4+\varepsilon}, \] uniformly in \(s\). As a corollary it follows that \(\Delta(n)\) has \(\gg_{\varepsilon}N^{1/4-\varepsilon}\) distinct square-free parts \(s\) as \(n\) ranges over the positive integers up to \(N\). It is also shown that \[ \sum_{s\leq S}T(N,s)\ll_{\varepsilon} N^{3/4+\varepsilon}S^{7/8}, \] whence one see that the square-free part of \(\Delta(n)\) is at least \(n^{2/7-\varepsilon}\) for ``almost-all'' \(n\). The proofs are based on the reviewer's ``square-sieve'' [Math. Ann. 266, 251--259 (1984; Zbl 0514.10038)]. This requires suitable estimates for character sums related to \(\Delta(n)\). Thus it is shown that if \(p,q\in(Q,2Q]\) are distinct primes, then \[ \sum_{n\leq N}\left(\frac{\Delta(n)}{pq}\right)\ll_{\varepsilon} \{NQ^{-1}\text{gcd}(p-1,q-1)+Q^3\}Q^{\varepsilon}. \]