2020-03-12, 02:50 | #1 |
"Sam"
Nov 2016
2^{2}·3·5^{2} Posts |
Orders of consecutive elements does not exceed floor(sqrt(p))
For some prime p, let m = ord_{p}(2) be the multiplicative order of 2 mod p, and m_{2} = ord_{p}(3) be the order of 3 mod p. Let L be the least common multiple of m and m_{2} (L = lcm(m,m_{2})).
Does a prime p exist such that L < sqrt(p) or simply floor(sqrt(p)) ? (There is no such prime below 10^9) The question in general is, for integers (a,b) (a ≠ b^{i} for some i > 2 or vice versa) are there finitely many primes p such that: L > floor(sqrt(p)) where L = lcm(m,m_{2}) m = ord_{p}(a) and m_{2} = ord_{p}(b) ? Last fiddled with by carpetpool on 2020-03-12 at 02:53 |
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