Modulo operation on large integer using -1

Question

In my textbook, there's an example of calculating the remainder of a large integer by using $-1$ like this:

$$4^{11}5^{13} \bmod 21 \equiv 20^{11}25 \bmod 21 \equiv (-1)^{11}4 \bmod 21 \equiv -4 \equiv 17 \,(\bmod 21)$$

It is not explained at any point. What am I missing?

Answer 1

It's simply $$\begin{align*}4^{11}5^{13}=4^{11}5^{11}5^2=(4\cdot5)^{11}\cdot5^2=(\color{red}{21}-1)^{11}\cdot 25&\equiv (\color{red}{0}-1)^{11}\cdot 25\pmod {\color{red}{21}}\\&\equiv {-1}^{11}\cdot 4\equiv...\pmod {\color{red}{21}}\end{align*}$$

Answer 2

This is because $4^{11} 5^{13} = 4^{11} \cdot 5^{11} \cdot 5^{2} = {4 \cdot 5}^{11} \cdot 25 = 20^{11} \cdot 25$

Once you have this, you use the fact that you are in $\mathbb{Z}/21\mathbb{Z}$ and the result follows since $20 = -1$ mod $21$ and $25 = 4$ mod $21$.

Answer 3

All it means is, in modular arithmetic, we generally get to pick from any representative of an equivalence class (an object ( in this case a number), that represents all equivalent objects(values), under an equivalence relation (a very useful type of relation)).

In this case it's, that two numbers have equivalent remainders on division by a certain number, because their difference is a multiple of that number.

Another example in modular arithmetic, is the Polynomial remainder theorem. It notes, in effect, that in the expansion of the binomial power $((x-a)+a)^n$ , that all terms except the last have a divisor of $x-a$ , meaning the whole expansions remainder on division by $x-a$ , is literally that last term, which is $a^n$ ; $x=(x-a)+a$ Though, so it applies to any polynomial in x termwise.