<s>[INST] Help proving Jacobi Symbol property $\left(\frac{-1}{n}\right) = (-1)^{\frac{n-1}{2}}$ Using the properties of Legendre symbols, I need to show $\left(\frac{-1}{n}\right) = (-1)^{\frac{n-1}{2}}$
Here is what I have so far
$\left(\frac{-1}{n}\right) = \left(\frac{-1}{p_1}\right)\left(\frac{-1}{p_2}\right) \cdots \left(\frac{-1}{p_k}\right)$ Where $p_1p_2\cdots p_k$ is the prime factorization of $n$. via the definition of Jacobi Symbols.
From the properties of Legendre symbols this is
$(-1)^{\frac{p_1 - 1}{2}}(-1)^{\frac{p_2 -1 }{2}} \cdots (-1)^{\frac{p_k-2}{2}} = (-1)^{\frac{p_1-1}{2}+\frac{p_2-1}{2}+\cdots + \frac{p_k-1}{2}}$
If I can show that $\frac{n-1}{2} \equiv \frac{p_1-1}{2}+\frac{p_2-1}{2}+\cdots + \frac{p_k-1}{2} \pmod{2}$. Then I would be there (I think).
I've been struggling with this for about an hour, and then decided to search. I found this question. However I don't understand the Lemma the answerer used in his proof.

Lemma 1
Let $a, b$ be odd integers.
Then $(ab - 1)/2 \equiv (a - 1)/2 + (b - 1)/2$ (mod $2$).
Proof:
Since $a - 1$ and $b - 1$ are even,
$(a - 1)(b - 1) \equiv 0$ (mod $4$).
Hence $ab - a - b + 1 \equiv 0$ (mod $4$).
Hence $ab - 1 \equiv (a - 1) + (b - 1)$ (mod $4$).
Hence $(ab - 1)/2 \equiv (a - 1)/2 + (b - 1)/2$ (mod $2$).

Specifically how we go from this (which I understand)

Hence $ab - a - b + 1 \equiv 0$ (mod $4$).

To this

Hence $ab - 1 \equiv (a - 1) + (b - 1)$ (mod $4$).

Which I'm sure really obvious, but I just cant see it :(
If I understood that, can I then just use it to finish off the proof by stating that
$\frac{p_1-1}{2}+\frac{p_2-1}{2}+\cdots + \frac{p_k-1}{2} \equiv \frac{n-1}{2}\pmod{2}$ via lemma above, and so it follows that
$\left(\frac{-1}{n}\right) = (-1)^{\frac{n-1}{2}}$
Thanks for any help or insight
 [/INST] You are making hard work of this. The point is to prove that
$$\left(\frac{-1}n\right)=\chi(n)\tag1$$
where
$$\chi(n)=\begin{cases}1&\text{if }n\equiv1\pmod 4,\\
-1&\text{if }n\equiv-1\pmod 4\end{cases}$$
for odd $n$. Now $\chi(mn)=\chi(m)\chi(n)$ and
also $\left(\frac{-1}{mn}\right)=\left(\frac{-1}m\right)
\left(\frac{-1}n\right)$. So if ($1$) holds for $m$ and $n$
it holds for $mn$, and it holds for all odd primes....
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