I've checked the factorization of $2^N – 1$ up through N = 120 for the largest prime factor, and it looks like the largest value of N where $2^N-1$ has a largest prime factor under 2500 is N = 60 (largest prime factor = 1321). As N gets larger, the largest prime factors get larger, even for the "abundant" numbers like 96, 108, and 120.
Is there a way to prove that no value of N > 60 exists such that the largest prime factor of $2^N – 1$ is less than 2500?
Best Answer
Zsigmondy's theorem implies that when $n \ge 7$, $2^n - 1$ has a prime divisor not dividing $2^k - 1$ for any $k < n$. So pick $n_0$ large enough that every prime number less than $2500$ divides $2^k - 1$ for some $k \le n_0$, then take $n > n_0$. This proves the desired claim for sufficiently large $n$ and then it suffices to check finitely many cases.