modular arithmetic - Why can we exchange numbers when working with modulo expressions?

Please excuse me if the answer is obvious because I'm a beginner.

Why can we exchange numbers when working with modulo expressions?

For example:

$$4^2 \equiv (-1)^2 \pmod{5}$$

You may say the replacement between $4$ and $-1$ is justified because:

$$4\equiv -1 \pmod{5}$$

I understand that equality, when you divide $4$ by $5$ you get a remainder $4$ and if we subtract $5$ from that we get $-1$. But I still don't understand why we can replace $4$ with $-1$.

Furthermore if $a\equiv c \pmod{b}$ are we justified in replacing $a$ with $c$ in every occasion?

Answer

You need the function you are dealing with to preserve multiplication. In fancier language, that means it is a homomorphism from $(\mathbb{Z},\cdot)$ to $(\mathbb{Z}/n\mathbb{Z},\cdot)$. In simpler language, that means that if $x,y$ are integers then $f(x \cdot y)=[x] \cdot [y]$, where the first $\cdot$ is integer multiplication, $[z]$ denotes the equivalence class of $z$ mod $n$, and the second $\cdot$ represents multiplication mod $n$. (Note that we often represent $[z]$ by the remainder of $z$ after division by $n$.)

For example $f : \mathbb{Z} \to \mathbb{Z}/n\mathbb{Z},f(x)=[x^2]$ is such a homomorphism, so $a^2 \equiv b^2 \mod n$ whenever $a \equiv b \mod n$. (Here $[y]$ denotes the equivalence class of $y$ mod $n$.) On the other hand, although $4 \equiv 9 \mod 5$, $2^4$ and $2^9$ are not equivalent mod 5.