# Jacobi Symbol

• Mar 7th 2010, 12:39 AM
kingwinner
Jacobi Symbol
Evaluate the Jacobi symbol ((n−1)(n+1)/n) for any odd natural number n.

Trying out some numbers, I THINK it alternates between 1 and -1, but how can we PROVE it formally?

Any help is appreciated!

[also under discussion in math link forum]
• Mar 7th 2010, 02:43 PM
chiph588@
$\displaystyle \left(\frac{(n-1)(n+1)}{n}\right) = \left(\frac{n-1}{n}\right)\left(\frac{n+1}{n}\right) = \left(\frac{-1}{n}\right)\left(\frac{1}{n}\right) = \left(\frac{-1}{n}\right)$

For $\displaystyle n$ odd, $\displaystyle \left(\frac{-1}{n}\right) = (-1)^\frac{n-1}{2} = \begin{cases} \;\;\,1 & \text{if }n \equiv 1 \pmod 4\\ -1 &\text{if }n \equiv 3 \pmod 4\end{cases}$.

For $\displaystyle n$ even, take $\displaystyle n=2k$, then $\displaystyle \left(\frac{-1}{n}\right) = \left(\frac{-1}{2k}\right) = \left(\frac{-1}{2}\right)\left(\frac{-1}{k}\right) = \left(\frac{-1}{k}\right) = \begin{cases} \;\;\,1 & \text{if }k \equiv 1 \pmod 4\\ -1 &\text{if }k \equiv 3 \pmod 4\end{cases}$.
• Mar 7th 2010, 04:27 PM
kingwinner
Quote:

Originally Posted by chiph588@
$\displaystyle \left(\frac{(n-1)(n+1)}{n}\right) = \left(\frac{n-1}{n}\right)\left(\frac{n+1}{n}\right) = \left(\frac{-1}{n}\right)\left(\frac{1}{n}\right) = \left(\frac{-1}{n}\right)$

For $\displaystyle n$ odd, $\displaystyle \left(\frac{-1}{n}\right) = (-1)^\frac{n-1}{2} = \begin{cases} \;\;\,1 & \text{if }n \equiv 1 \pmod 4\\ -1 &\text{if }n \equiv 3 \pmod 4\end{cases}$.

For $\displaystyle n$ even, take $\displaystyle n=2k$, then $\displaystyle \left(\frac{-1}{n}\right) = \left(\frac{-1}{2k}\right) = \left(\frac{-1}{2}\right)\left(\frac{-1}{k}\right) = \left(\frac{-1}{k}\right) = \begin{cases} \;\;\,1 & \text{if }k \equiv 1 \pmod 4\\ -1 &\text{if }k \equiv 3 \pmod 4\end{cases}$.

Thank you!

But I thought the Jacobi symbol is defined only for ODD positive integers at the bottom. In your proof, why is there a case where "n" is even??? How is this possible??
• Mar 7th 2010, 06:42 PM
chiph588@
Quote:

Originally Posted by kingwinner
Thank you!

But I thought the Jacobi symbol is defined only for ODD positive integers at the bottom. In your proof, why is there a case where "n" is even??? How is this possible??

Oops! You're right!
• Mar 7th 2010, 07:24 PM
kingwinner
But I'm concerned with one special case. For the case n=1, ((n−1)(n+1)/n)=(0/1)

Is (0/1)=1 or (0/1)=0 ?? Which one is the correct answer and why?

Thanks!
• Mar 7th 2010, 09:33 PM
chiph588@
Quote:

Originally Posted by kingwinner
But I'm concerned with one special case. For the case n=1, ((n−1)(n+1)/n)=(0/1)

Is (0/1)=1 or (0/1)=0 ?? Which one is the correct answer and why?

Thanks!

$\displaystyle \left(\frac{a}{n}\right) = \begin{cases} \;\;\,0\mbox{ if } \gcd(a,n) \ne 1 \\\pm1\mbox{ if } \gcd(a,n) = 1\end{cases}$

So sub in $\displaystyle 0$ and $\displaystyle 1$ and see what you get.
• Mar 7th 2010, 09:40 PM
kingwinner
Quote:

Originally Posted by chiph588@
$\displaystyle \left(\frac{a}{n}\right) = \begin{cases} \;\;\,0\mbox{ if } \gcd(a,n) \ne 1 \\\pm1\mbox{ if } \gcd(a,n) = 1\end{cases}$

So sub in $\displaystyle 0$ and $\displaystyle 1$ and see what you get.

gcd(0,1)=1, so that rule says that (0/1)=+1 OR -1, but how do we know whether it is +1 or -1?
• Mar 7th 2010, 09:46 PM
chiph588@
Jacobi symbol

Apparently the answer is $\displaystyle 0$. I haven't had too much exposure to the Jacobi symbol so I can't really tell you why. Check out the site for yourself though.
• Mar 7th 2010, 10:18 PM
NonCommAlg
$\displaystyle \left( \frac{0}{1} \right)=1$ because the set of prime factors of $\displaystyle 1$ is empty. if $\displaystyle n > 1$ is an odd integer, then $\displaystyle \left( \frac{0}{n} \right)=0.$
• Mar 8th 2010, 12:08 AM
kingwinner
Quote:

Originally Posted by NonCommAlg
$\displaystyle \left( \frac{0}{1} \right)=1$ because the set of prime factors of $\displaystyle 1$ is empty. if $\displaystyle n > 1$ is an odd integer, then $\displaystyle \left( \frac{0}{n} \right)=0.$

Is there any reason why (0/1)=1?? Is this simply becuase by convention, we define it to be that way?
1 has no prime factorization, so the product is empty. Is it conventional to define the "empty" product to be equal to +1??

Also, is it true that, by definition, (a/1)=1 for any integer a?

Can someone clarify this? Thank you!
• Mar 8th 2010, 06:54 PM
NonCommAlg
Quote:

Originally Posted by kingwinner
Is there any reason why (0/1)=1?? Is this simply becuase by convention, we define it to be that way?
1 has no prime factorization, so the product is empty. Is it conventional to define the "empty" product to be equal to +1??

Also, is it true that, by definition, (a/1)=1 for any integer a?

Can someone clarify this? Thank you!

• Mar 8th 2010, 07:52 PM
kingwinner
Quote:

Originally Posted by NonCommAlg

Is there any real reason why we define (a/1)=1 for any integer a?

Why not define it to be 0? why not -1?

• Mar 8th 2010, 08:53 PM
Bacterius
Quote:

Originally Posted by kingwinner
Is there any real reason why we define (a/1)=1 for any integer a?

Why not define it to be 0? why not -1?