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]

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- Mar 7th 2010, 12:39 AMkingwinnerJacobi 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 PMchiph588@
$\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 PMkingwinner
- Mar 7th 2010, 06:42 PMchiph588@
- Mar 7th 2010, 07:24 PMkingwinner
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 PMchiph588@
- Mar 7th 2010, 09:40 PMkingwinner
- Mar 7th 2010, 09:46 PMchiph588@
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 PMNonCommAlg
$\displaystyle \left( \frac{0}{1} \right)=1$ 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 AMkingwinner
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 PMNonCommAlg
- Mar 8th 2010, 07:52 PMkingwinner
- Mar 8th 2010, 08:53 PMBacterius
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 :)