Thursday, 23 January 2014

elementary number theory - Solving a congruence, tricky implication



We want to prove that



243x1mod2018x4033mod2018



My try :



Assume that 243x1mod2018




We have x20161mod2018 (by Fermat (1009 is prime) and oddness of x) so x2015243mod2018



but 403×5=2015



hence (x403)5243mod2018 or equivalently (x403)535mod2018. I wonder how to get from this last congruence to the desired x4033mod2018 ?



Thanks for any suggestions.


Answer



From the last step of your attempt i.e. (x403)535mod2018, you can go to the conclusion if you can show that y5243mod2018 has only one solution mod 2018 (because this would force x4033mod2018 as they are both solutions).




While that statement is true, its proof is not going to be easy because it boils down to asking why x51mod2018 has a unique solution, and this is not clear at all.



Therefore, you have not done anything wrong but got yourself in a higher power than required.



However, you did do some groundwork. For example, by noting that 243x1mod2018, so we may raise both sides to the power 403 :
35x1mod201832015x4031mod2018



Now, multiply by a further 3, and eliminate the 32016 using an argument made in your attempt.



No comments:

Post a Comment

real analysis - How to find limhrightarrow0fracsin(ha)h

How to find limh0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...