DIVISIBILITY CONGRUENCE question

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Question: If gcd(a,42)=1, show that a^6 - 1 is divisible by 168.

Answer: So I know that if 42 were prime, than the Little Fermat Thm says that a^p-1 is congruent to 1 mod p. But I have no idea where to start if p is not prime. Help please.
 
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prove the divisibility by 3, 7 and 8 separately. 3 and 7 can be done with Little Fermat's theorem.
For divisibility by 8, factor a^6 -1.
 

Thread 'Trouble understanding an online solution to an exercise in Dummit & Foote'

##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
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Thread 'Questions about non existence of GCDs vs (coimages, cokernels)'

The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
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Thread '##(A/\mathfrak{a})_{\mathfrak{p}/\mathfrak{a}}## and its isomorphism?'

I asked online questions about Proposition 2.1.1: The answer I got is the following: I have some questions about the answer I got. When the person answering says: ##1.## Is the map ##\mathfrak{q}\mapsto \mathfrak{q} A _\mathfrak{p}## from ##A\setminus \mathfrak{p}\to A_\mathfrak{p}##? But I don't understand what the author meant for the rest of the sentence in mathematical notation: ##2.## In the next statement where the author says: How is ##A\to...
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