A note on the prime factors of Fermat numbers (Q1769324)

Let \(F_m=2^{2^m}+1\) be the \(m\)th Fermat number, and \(F_m=p_1p_2\cdots p_l\) with primes \(p_1\leq p_2\leq\cdots\leq p_l\). For any positive integer \(k\), let \(P(k)\) be the largest prime factor \(k\). In this setting, \textit{M. Le} [Southeast Asian Bull. Math. 22, No. 1, 41--44 (1998; Zbl 0937.11001)] proved that \(P(F_m)>2^{m-4}m\) for \(m\geq 2^{18}\). \textit{A. Grytczuk, M. Wójtowicz} and \textit{F. Luca} [Southeast Asian Bull. Math. 25, No. 1, 111--115 (2001; Zbl 1009.11005)] improved this result to \(P(F_m)\geq 2^{m+2}(4m+9)+1\) for \(m\geq 4\). In this paper, the author gives a further improvement of this inequality to \[ P(F_m)> 2^{m+2}(4m+16+\log_2(m+7))+1\text{ if }m\geq 4, \] where \(\log_2\) denotes the base 2 logarithm. The proof is elementary.