Using modular arithmetic in exponent

Question

For odd positive integers $n_1$ and $n_2$ I am giver a relation:
$\frac{n_1n_2-1}{2}\equiv \frac{n_1-1}{2} + \frac{n_2-1}{2} \pmod{2}$

So I'm proving something and on some step of my prove I want to use giver relation and write something like this:
$(-1)^{(n_1-1)/2 + (n_2-1)/2 }= (-1)^{(n_1n_2-1)/2}$

Can I do this? And what do I need to claim if I want to use giver relation like this?

Answer 1

Yes you can. From definition, for some $k\in\mathbb Z$:

$$\frac {n_1n_2-1}2+2k=\frac{n_1-1}2+\frac{n_2-1}2$$

Hence:

$$(-1)^{(n_1-1)/2+(n_2-1)/2}=(-1)^{2k+(n_1n_2-1)/2} = (-1)^{2k}(-1)^{(n_1n_2-1)/2}=(-1)^{(n_1n_2-1)/2}$$

Answer 2

Hint: Prove that

$$(-1)^a \equiv (-1)^b \Leftrightarrow a\equiv b \pmod{2}.$$