The discussion revolves around proving the congruence \( n^{21} \equiv n \mod 30 \) for an integer \( n \). Participants reference related congruences involving \( n^7 \equiv n \mod 42 \) and \( n^{13} \equiv n \mod 2730 \) as part of their reasoning.
Several participants are actively engaging with the problem, providing insights into the divisibility by \( 2 \) and \( 3 \). There is a sense of progression as they build on each other's ideas, but no consensus has been reached yet regarding the next steps for proving the necessary conditions.
Participants are working under the constraints of proving the congruence without providing complete solutions, focusing on reasoning and exploration of assumptions related to the problem.
yeland404 said:Homework Statement
let n be an integer . Prove the congruence below.
n^21 \equiv n mod 30
Homework Equations
n^7 \equiv n mod 42
n^13 \equiv n mod 2730
The Attempt at a Solution
to prove 30| n^21-n，it suffices to show 2|n^21-n,3|n^21-n,5|n^21-n
and how to prove them?
yeland404 said:then n^21-n = n(n^20-1), suppose n is even , then 2|n^21-n
if n is odd, n^20 is odd, so n^20-1 is even;
to 3, it means n^21=(n^3）^7=n^7=(n^3)^2*n
then how is the next to prove 3|n(n^20-1)
Dick said:You are almost there with this line, "to 3, it means n^21=(n^3）^7=n^7=(n^3)^2*n". Think about it a little more and you will get it.