In the ring $\mathbb Z_5$ (class of modulo 5) compute $[3]^{-4}$.

Question

$[3]$ means $x \equiv 3\mod(5)$. I know how to compute $[3]^{4}$ that is positive powers but I have not been in a situation where I need to take something modulo(n) to a negative power. the book says $[3]^{-1}=[2]$ but why is this? what are the steps I need to take to find this?

Answer 1

When doint arithmetic modulo $\;p\;$ , you get that $\;x^{-1}=a\pmod p\iff xa=1\pmod p\;$ . In this case,

$$3^{-1}=2\pmod 5\;\;\text{since}\;\;3\cdot2=6=1\pmod 5$$

You can do the same modulo any number $\;p\;$ ...under the condition, of course, that you have an invertible number. If the modulo is prime then any non-multiple of that prime is invertible.

Answer 2

By little Fermat $\,3^{-4}\equiv (3^{-1})^4\equiv 1\,$ by $\,3^{-1}\not\equiv 0\ $ (or brute-force: $\ (3^{-1})^4\!\equiv 2^4\equiv 1)$

Alternatively use $\ \underbrace{3^{4}}_{\large\equiv\ 1}\, 3^{-4}\equiv 1$

Answer 3

By definition $3^{-4}$ is the equivalence class $x$ where $x\cdot3^4 \equiv 1 \pmod 5$ (if any).

We could calculate that $3^4 =81\equiv 1 \pmod 5$ and to so $x\cdot 3^4 \equiv x \cdot 1\equiv x \equiv 1 \pmod 5$ gives us ... $x \equiv 1 \pmod 5$.

So $3^{-4}\equiv 1 \pmod 5$

But that'd pretty ineffectual and ignores many things we should have at our finger tips.

If $p$ is prime $Z_p$ is a field but if not then $Z_p$ has some zero divisors.

If $a^{-1}$ so that $a^{-1}\cdot a \equiv 1$ then $a^{-k} = (a^{-1})^k$ so we could attempt to find $3^{-1}$ by solve $3a\equiv 1\pmod 5$ via $3a \equiv 6\pmod 5$ so (as $3,5$ are relatively prime) then $3^{-1}= a \equiv 2 \pmod 5$ and $3^{-4}\equiv (3^{-1})^4 \equiv 2^4\equiv 16\equiv 1$.

But most importantly, we were ignoring that $Z_5$ is a fields so $3^{-1}$ exists and we don't need to figure out what it is, it's enough to no it exists and that by Fermat's Little Theorem for all $a \not\equiv 0$ $a^4 \equiv 1 \pmod 5$ so we can figure either:

$3^{-4}\equiv 3^{4-4} \equiv 3^0 \equiv 1 \pmod 5$ or

$3^{-4}\equiv (3^{-1})^4 \equiv 1 \pmod 5$.