Arithmetic with congruence classes

Question

I need to compute the following expression in $\mathbb{Z}_{5, +, \cdot}$ : $$ [2]_5^4 - [4]_5^4 \cdot [3]_5^4 \cdot [2]_5^4 $$ I'm not sure what is the best way to do this. Should I determine all the terms separately first? For example, for the second term $$ 4^1 = 4 \\ 4^2 = 16 = 1 \\ 4^3 = 4 \cdot 1 = 4 \\ 4^4 = 4 \cdot 4 = 1 $$ etc. So this means $[4]_5^4 = [1]_5$ ? Also, I'm not sure if I can use Fermat little theorem here, which says that if $p$ is prime and doesn't divide $a$, then $[1]_p = [a^{p-1}]_p$. But can I say that $[a^{p-1}]_p = [a]_p^{p-1} $ or is this false?

Answer 1

Yes, $[a^{p-1}]_p = [a]_p^{p-1}$ because $[xy]_p = [x]_p[y]_p$ for all $x, y \in \Bbb{Z}_5$.

You're on the right track but I wouldn't say it like "$4^2 = 1$" because that's technically incorrect. It's better to use modular equivalences, as in "$4^2 \equiv 1 \pmod 5$."

Let me know if you need more assistance.

Answer 2

Hint:

Use Lil' Fermat: for any $a\not\equiv0\mod 5$, one has $\;a^4\equiv 1\mod 5$.