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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

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  • $\begingroup$ In $\left(\frac{-1}{n}\right) = (-1)^{\frac{n-1}{2}}$ the LHS is completely multipliccative by definition. Show the RHS is completely multiplicative too, thus it suffices to look at $n$ prime. $\endgroup$
    – reuns
    Commented Oct 23, 2017 at 3:12
  • $\begingroup$ @reuns I'm not sure I follow....show $(n-1) = ( (p_1p_2\cdots p_k) -1)$? $\endgroup$ Commented Oct 23, 2017 at 3:38
  • $\begingroup$ For $n$ odd let $f(n) = \left(\frac{-1}{n}\right)$ and $g(n) = (-1)^{\frac{n-1}{2}}$, for $n$ even set $f(n) = g(n)=0$. By definition $f(n) = \prod_{p^k \| n} f(p)^k$. And it is clear that $g(mn) = g(m)g(n)$ so $g(n) = \prod_{p^k \| n} g(p)^k$. Thus it suffices to check that $f(p) = g(p)$. $\endgroup$
    – reuns
    Commented Oct 23, 2017 at 4:06

1 Answer 1

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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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  • $\begingroup$ that makes sense, and I think I understand it, but if I don't explicitly tie $\frac{n-1}{2}$ into it, I wont get any credit. or is that obvious in there and I just dont see that either, lol. Thanks for your help $\endgroup$ Commented Oct 23, 2017 at 3:25
  • $\begingroup$ $\chi(n)$ is $(-1)^{(n-1)/2}$ $\endgroup$ Commented Oct 23, 2017 at 5:59

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