Determine if $f: \Bbb Z_{4711} \rightarrow \Bbb Z_{4711}, f(x) = 1632x+1792$ is injective, surjective, bijective and/or invertible

Question

I have no idea how to get started with such a large $\Bbb Z_{4711}$. The solutions say

"Since $\gcd(1632,4711)=1$ that means 1632 in invertible in $\Bbb Z_{4711}$. So f is invertible and is $f^{-1}(x)=1632^{-1}(x-1792)$ and hence is bijective, injective and surjective."

The definitions of invertibility that we were given referred to numbers, not functions:

A number $a \in \Bbb Z_m$ is invertible iff $gcd(a,m)=1$ (a,m coprime).

and

A number $a \in \Bbb Z_m$" is invertible if $\exists x \in \Bbb Z_m$ such that $ax\equiv 1 \pmod m$. x is an in verse to a.

I don't understand why the solution is interested in checking whether the number 1632 is invertible in $\Bbb Z_{4711}$?

Answer 1

You wrote

I don't understand why the solution is interested in checking

why the number 1632 is invertible in $\Bbb Z_{4711}$

Say $y\equiv 1632x+1792$ and we wish to solve for $x$ in terms of $y$.

Well, $y-1792\equiv 1632x$, and then $1632$ is invertible modulo $4711$

(that is, there is an element $k$ of $\mathbb Z_{4711}$ such that $1632k\equiv1\bmod4711$)

implies we can multiply both sides by $k$ to solve for $x$: $$k(y-1792)\equiv x.$$

To illustrate the problems that can occur if the number is not invertible,

consider $y-1\equiv2x$ and $y-2\equiv2x\bmod 4;$

$2$ is not invertible $\bmod 4$,

so there may not be a unique solution $x$.