Divisors of $-1$ are only $1$ and $-1$? I'm working through a discrete math textbook and I've come across this question with answer:
Prove that the only divisors of $−1$ are $−1$ and $1$.
Answer:
We established that $1$ divides any number; hence, it divides $−1$, and any nonzero number divides itself. Thus, $1$ and $−1$ are divisors of $−1$. To show that these are the only ones, we take $d$, a positive divisor of $−1$. Thus, $dk = −1$ for some integer $k$, and $(−1)dk = d(−k) = (−1)(−1) = 1$; hence, $d\mid 1$, and the only divisors of $1$ are $1$ and $−1$. Hence, $d = 1$ or $d = −1$.
I understand everything stated in the answer except for the part: $(-1)dk = d(-k) = (-1)(-1) = 1$
Perhaps someone could help me understand where the $(-1)dk$ comes from? And how we go from $d(-k)$ to $(-1)(-1)$?

We are assuming that $d$ is a divisor of $-1$ that is
$$dk=-1$$
and multiplying each side by $-1$ we obtain
$$-1\cdot dk=-1\cdot (-1)=1 \iff d(-k)=1 \iff d=1,-1 $$
