We want to prove that
243x≡1mod2018⟹x403≡3mod2018
My try :
Assume that 243x≡1mod2018
We have x2016≡1mod2018 (by Fermat (1009 is prime) and oddness of x) so x2015≡243mod2018
but 403×5=2015
hence (x403)5≡243mod2018 or equivalently (x403)5≡35mod2018. I wonder how to get from this last congruence to the desired x403≡3mod2018 ?
Thanks for any suggestions.
Answer
From the last step of your attempt i.e. (x403)5≡35mod2018, you can go to the conclusion if you can show that y5≡243mod2018 has only one solution mod 2018 (because this would force x403≡3mod2018 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 x5≡1mod2018 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 243x≡1mod2018, so we may raise both sides to the power 403 :
35x≡1mod2018⟹32015x403≡1mod2018
Now, multiply by a further 3, and eliminate the 32016 using an argument made in your attempt.
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