# Thread: Jacobi Symbol Properties

1. ## Jacobi Symbol Properties

This is a theorem about Jacobi symbols in my textbook:
Let n and m be ODD and positive. Then (a/nm)=(a/n)(a/m) and (ab/n)=(a/n)(b/n)
Moreover,
(i) If gcd(a,n)=1, then ($\displaystyle a^2/n$) = 1 = ($\displaystyle a/n^2$)
(ii) If gcd(ab,nm)=1, then ($\displaystyle ab^2/nm^2$)=(a/n)
=====================================

(i) is easy and follows from the definition, but how can we prove (ii)? My textbook stated the theorem without proof and just says the proofs are easy, but I have no idea why (ii) is true.

Any help is appreciated!

2. Originally Posted by kingwinner
This is a theorem about Jacobi symbols in my textbook:
Let n and m be ODD and positive. Then (a/nm)=(a/n)(a/m) and (ab/n)=(a/n)(b/n)
Moreover,
(i) If gcd(a,n)=1, then ($\displaystyle a^2/n$) = 1 = ($\displaystyle a/n^2$)
(ii) If gcd(ab,nm)=1, then ($\displaystyle ab^2/nm^2$)=(a/n)
=====================================

(i) is easy and follows from the definition, but how can we prove (ii)? My textbook stated the theorem without proof and just says the proofs are easy, but I have no idea why (ii) is true.

Any help is appreciated!
Think it is because...

$\displaystyle \bigg{(}\frac{ab^2}{nm^2}\bigg{)} = \bigg{(}\frac{a}{nm^2}\bigg{)}\bigg{(}\frac{b^2}{n m^2}\bigg{)} = \bigg{(}\frac{a}{nm^2}\bigg{)}\cdot 1$

$\displaystyle = \bigg{(}\frac{a}{n}\bigg{)}\bigg{(}\frac{a}{m^2}\b igg{)} = \bigg{(}\frac{a}{n}\bigg{)}\cdot 1 = \bigg{(}\frac{a}{n}\bigg{)}$

Was it supposed to be $\displaystyle \bigg{(}\frac{(ab)^2}{(nm)^2}\bigg{)}$? That was a bit unclear

3. Originally Posted by Deadstar
Think it is because...

$\displaystyle \bigg{(}\frac{ab^2}{nm^2}\bigg{)} = \bigg{(}\frac{a}{nm^2}\bigg{)}\bigg{(}\frac{b^2}{n m^2}\bigg{)} = \bigg{(}\frac{a}{nm^2}\bigg{)}\cdot 1$

$\displaystyle = \bigg{(}\frac{a}{n}\bigg{)}\bigg{(}\frac{a}{m^2}\b igg{)} = \bigg{(}\frac{a}{n}\bigg{)}\cdot 1 = \bigg{(}\frac{a}{n}\bigg{)}$
Why (b^2/nm^2)=1? How do you know that gcd(b,nm^2)=1?
Why (a/m^2)=1? How do you know that gcd(a,m)=1?

Was it supposed to be $\displaystyle \bigg{(}\frac{(ab)^2}{(nm)^2}\bigg{)}$?
No.

Thanks for explaining!

4. Originally Posted by kingwinner
Why (b^2/nm^2)=1? How do you know that gcd(b,nm^2)=1?
Why (a/m^2)=1? How do you know that gcd(a,m)=1?
Since $\displaystyle \left(\frac{xy}{r}\right) = \left(\frac{x}{r}\right)\left(\frac{y}{r}\right)$, we have that $\displaystyle \left(\frac{b^2}{nm^2}\right) = \left(\frac{b}{nm^2}\right)\left(\frac{b}{nm^2}\ri ght) = \left(\frac{b}{nm^2}\right)^2 =1$

The other case follows similarly.