Checking my work for $21^{100}-12^{100}$ modulo $11$ 
Question 1: Is $21^{100}-12^{100}$ divisible by $11$?

My work: Note:$$\begin{align*} & \color{red}{21^{100}}\equiv10^{100}\equiv 100^{50}\equiv1^{50}\equiv1 \mod 11\\ & \color{blue}{12^{100}}\equiv 1^{100}\equiv 1\mod 11\end{align*}$$
Hence,$$\begin{align*}\color{red}{21^{100}}-\color{blue}{12^{100}}\equiv 1-1\equiv 0\mod 11\end{align*}$$
Hence, $21^{100}-12^{100}$ is divisible by $11$.


Actual Question:



*

*Does my work hold? Is it logical and reasonable?

*Is there a quicker way of solving this problem?


I am very new to modular arithematic (i.e I just learned it $10$ minutes ago) so tips on a simple book where I can learn a bit more in-depth stuff about modular arithematics greatly helps!
 A: An easier approach for such problems is making use of Fermat's_little_theorem
which says, for any integer $a$, 
$a^{p-1}\equiv 1 \mod{p}$
if $p$ is prime and $gcd(a,p)=1$
Using this, we can solve the problem as follows.
$21^{11-1}\equiv 1 \pmod {11}\\
21^{10}\equiv 1 \pmod {11}\\
21^{100}\equiv 1 \pmod {11} \cdots(1)$
$12^{11-1}\equiv 1 \pmod {11}\\
12^{10}\equiv 1 \pmod {11}\\
12^{100}\equiv 1 \pmod {11} \cdots(2)$
Therefore $21^{100}-12^{100} \equiv 0 \pmod {11}$
A: Your calculation looks correct. It would be more conventional (and slightly less work) to write it as:
$$ 21\equiv -1 \pmod{11} \qquad\text{and}\qquad 12\equiv 1 \pmod{11} $$
so
$$ 21^{100} - 12^{100} \equiv (-1)^{100} - 1^{100} \equiv 1 - 1 \equiv 0 \pmod{11} $$
since $-1$ to an even power is always $1$.
A: $$\begin{align}
21^{100} - 12^{100} & \equiv (11 \times 2 - 1)^{100} - (11 + 1)^{100} \\
& \equiv (0 \times 2 - 1)^{100} - (0 + 1)^{100} \\
& \equiv (-1)^{100} - 1^{100} \\
& \equiv 1 - 1 \\
& \equiv 0 \pmod {11}
\end{align}$$
A: $21^{100}-12^{100}\equiv21^0-12^0\equiv0($mod $11)$
since
$21^{2k}\equiv12^{k}\equiv1($mod $11)$
