3 ms·
Also usefully, (2(a+1)^n-1) % a = (2a(a+1)^(n-1) + 2(a+1)^(n-1) - 1) % a = 1. So if you choose one of the 3k+1 primes, it's also guaranteed not divisible by 3.
by naniwaduni 5y ago
Also usefully, (2(a+1)^n-1) % a = (2a(a+1)^(n-1) + 2(a+1)^(n-1) - 1) % a = 1. So if you choose one of the 3k+1 primes, it's also guaranteed not divisible by 3.