mersenneforum.org How to prove that only Powers of 6^(6+35j) are of the form648+23004s
 Register FAQ Search Today's Posts Mark Forums Read

 2022-09-14, 18:21 #1 enzocreti   Mar 2018 2·269 Posts How to prove that only Powers of 6^(6+35j) are of the form648+23004s This Is a Little more difficult: How to prove that Powers 6^(6+35j) for j nonnegative integer are of the form 648+23004s and only that powers . I mean there Is no orher 6^g of the form 648+23004s, only Powers 6^(6+35j) Last fiddled with by enzocreti on 2022-09-14 at 19:20
 2022-09-15, 13:49 #2 Dr Sardonicus     Feb 2017 Nowhere 137308 Posts If a congruence holds (mod n), it holds (mod d) for every divisor d of n. In particular, in the case at hand, as you more or less pointed out in this post, if 6^e == 648 (mod 23004), then 6^e == 648 (mod 71), or 6^e == 9 (mod 71). Since 71 is a prime, the values of e are easily characterized.

 Similar Threads Thread Thread Starter Forum Replies Last Post drkirkby Homework Help 14 2021-04-29 15:35 George M Miscellaneous Math 5 2018-01-02 11:11 PawnProver44 Miscellaneous Math 40 2016-03-19 07:33 nibble4bits Math 31 2007-12-11 12:56 Numbers Puzzles 3 2005-07-13 04:42

All times are UTC. The time now is 23:54.

Tue Dec 6 23:54:19 UTC 2022 up 110 days, 21:22, 0 users, load averages: 0.73, 0.90, 0.93