For context:
The question/answer is given as follows:
Show by induction that if n is a positive integer, then $4^n ≡ 1 + 3n (\text{mod } 9)$.
For the base case, $4 ≡ 1+3 (\text{mod } 9)$.
For the induction hypothesis, assume that $4n ≡ 1+3n (\text{mod } 9)$ for some positive integer n.
Then:
$4^{n+1} = 4^n \cdot 4 ≡ 4(1 + 3n) = 4 + 12n \equiv 4 + 3n \equiv 1 + 3(n + 1) (\text{mod } 9)$.
Therefore $4^n ≡ 1 + 3n (\text{mod } 9)$ for all positive integers $n$.
Question:
What property allows you to show that: $4 + 12n \equiv 4 + 3n \ (\text{mod } 9)$? I can plugin numbers and show, but this is not very constructive. I also note that:
$ 4+12n \equiv 4+3n \ (\text{mod } 9) \Rightarrow 9 | (4+12n)-(4+3n) \Rightarrow 9 | 9n $.
If I was not given $4+3n$ to begin with, I don't know if I could have concluded that result.
The taking of modulus respects multiplication and addition. So if $a\equiv b\bmod n$, then $a\cdot c\equiv b\cdot c\bmod n$ and $a+c\equiv b+c\bmod n$ for any integer $c$.
In your case, $12\equiv 3\bmod 9$, so $4+12n\equiv 4+3n\bmod 9$.