Is $11\mid21^{100} - 12^{100}$ true? Earlier on the book I was doing, it was established that $m\mid a_1^n-a_2^n$ if $m\mid a_1-a_2$. Because of this, I ignored the signs and said that since $21-12$ is not a multiple of 11, neither is $21^{100} - 12^{100}$. But I was wrong and I can't understand what I did wrong. Can anyone explain to me what I did wrong?
 A: The claim you are referring to is $$m|a_1-a_2 \implies m|a_1^n-a_2^n=(a_1-a_2)(\ldots)$$ but the converse is not necessarly true.
In this case we have that
$$21\equiv -1 \mod 11 \implies 21^{100}\equiv 1 \mod 11$$
and
$$12\equiv 1 \mod 11 \implies 12^{100}\equiv 1 \mod 11$$
therefore
$$21^{100} - 12^{100}\equiv 0 \mod 11$$
A: Your error is an error in logic. If $P \Longrightarrow Q$, it doesn't mean that $\neg P \Longrightarrow \neg Q$.
For example if it's snowing, it implies that the sky is grey, but if it's not snowing, it doesn't imply that the sky is not grey.
A: You have the implication the wrong direction... $a^n - b^n$ can be a multiple of $m$ even if $a - b$ is not. But if $a - b$ is a multiple of $m$ then $a^n - b^n$ has to also be a multiple of $m$. 
You can still use this method for the problem at hand. Let $a = 21^2 = 441$ and $b = 12^2 = 144$. Then $a - b = 297$ is equal to $11*27$, so is a multiple of $11$. Hence $a^{50} - b^{50} = 21^{100} - 12^{100}$ is also a multiple of $11$.
