Jacobi Symbol

**kingwinner** · March 7th 2010, 12:39 AM

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]

**chiph588@** · March 7th 2010, 02:43 PM

$\left(\frac{(n-1)(n+1)}{n}\right) = \left(\frac{n-1}{n}\right)\left(\frac{n+1}{n}\right) = \left(\frac{-1}{n}\right)\left(\frac{1}{n}\right) = \left(\frac{-1}{n}\right)$

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

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

**kingwinner** · March 7th 2010, 04:27 PM

Originally Posted by chiph588@

$\left(\frac{(n-1)(n+1)}{n}\right) = \left(\frac{n-1}{n}\right)\left(\frac{n+1}{n}\right) = \left(\frac{-1}{n}\right)\left(\frac{1}{n}\right) = \left(\frac{-1}{n}\right)$

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

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

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??

**chiph588@** · March 7th 2010, 06:42 PM

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!

**kingwinner** · March 7th 2010, 07:24 PM

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!

**chiph588@** · March 7th 2010, 09:33 PM

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!

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

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

**kingwinner** · March 7th 2010, 09:40 PM

Originally Posted by chiph588@

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

So sub in $0$ and $1$ 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?

**chiph588@** · March 7th 2010, 09:46 PM

Jacobi symbol

Apparently the answer is $0$ . 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.

**NonCommAlg** · March 7th 2010, 10:18 PM

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

**kingwinner** · March 8th 2010, 12:08 AM

Originally Posted by NonCommAlg

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

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!

**NonCommAlg** · March 8th 2010, 06:54 PM

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!

the answer is "yes" to both your questions.

**kingwinner** · March 8th 2010, 07:52 PM

Originally Posted by NonCommAlg

the answer is "yes" to both your questions.

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?

Thanks for answering!

**Bacterius** · March 8th 2010, 08:53 PM

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?

Thanks for answering!

It could perhaps be to avoid making annoying "special cases" in theorems.

Example (which is not linked to your question, but just to show you) :

Why do we define 1 as not being prime ?

If we defined 1 as being prime, we would have to reformulate the Fundamental Theorem of Arithmetic in a rather heavy way, stating that "apart from adding 1's, the prime decomposition of a number is unique without taking into account the order" ...

And since there is no particular reason to define 1 as a prime, we just prefer to say it isn't one

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