The square-free kernel of \(x^{2^n}-a^{2^n}\) (Q2773330)

Let \(t=2^n\). The author considers the number \(\nu(x^t-a^t)\) of odd prime factors of the square-free kernel of numbers \(x^t-a^t\), where \(x>a\geq 1\) and \(n\geq 2\). Using properties of quadratic sequences (or binary linearly recurring sequences) and a previous theorem of his [preprint, 2001], he proves that under a reasonable assumption, for each \(a\geq 1\) the set NEWLINENEWLINENEWLINE\(T_a= \{(x,n)\mid n\geq 2,\;x> a\) and the square-free kernel of \(x^t-a^t\) has \(n-1\) odd prime factors\} NEWLINENEWLINENEWLINEis finite and effectively computable. In the final section the author shows with several examples how to determine \(T_a\) for \(a=1,2,3,4,6,10\). He also shows the following results: NEWLINE\[NEWLINE\begin{aligned} \nu(3^t-1) &\geq n\quad\text{for }n\geq 4,\\ \nu(7^t-1) &\geq n\quad\text{for }n\geq 4,\\ \nu(99^t) &\geq n\quad\text{for }n\geq 3,\\ \text{and} \nu(x^t-1) &\geq n\quad\text{for }n\geq 2, \quad\text{for }x\neq 3,7,99. \end{aligned}NEWLINE\]NEWLINE{}.