# Thread:

1. ## 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]

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

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

3. 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??

4. 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!

5. 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!

6. 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!
$\left(\frac{a}{n}\right) <br /> = \begin{cases}<br /> \;\;\,0\mbox{ if } \gcd(a,n) \ne 1<br /> <br /> \\\pm1\mbox{ if } \gcd(a,n) = 1\end{cases}$

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

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

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?

8. 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.

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

10. 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!

11. 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.

12. 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!

13. 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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