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Suppose that $a,b \in \mathbb{N}$ are both odd and $m \in \mathbb{Z}.$ I want to show that

$$\left(\dfrac{m}{ab}\right) = \left(\dfrac{m}{a}\right) \left(\dfrac{m}{b}\right)$$

Here, we are talking about the Jacobi Symbol, hence why I put the $\ell$ subscript on each term. I have solved the case when

$$\left(\dfrac{ab}{m}\right) = \left(\dfrac{a}{m}\right) \left(\dfrac{b}{m}\right)$$

Here, we can let $m$ = $p_1 p_2...p_r,$ be the prime factorization of $m$ and we can re-write this as

$$\left(\dfrac{ab}{m}\right) = \left(\dfrac{ab}{p_1}\right) \left(\dfrac{ab}{p_2}\right)... \left(\dfrac{ab}{p_r}\right) = \left(\dfrac{a}{p_1}\right) \left(\dfrac{b}{p_1}\right)... \left(\dfrac{a}{p_r}\right) \left(\dfrac{b}{p_r}\right) = \left(\dfrac{a}{p_1}\right) \left(\dfrac{a}{p_2}\right)... \left(\dfrac{a}{p_r}\right) \left(\dfrac{b}{p_1}\right) \left(\dfrac{b}{p_2}\right)...\left(\dfrac{b}{p_r}\right) = \left(\dfrac{a}{m}\right)\left(\dfrac{b}{m}\right)$$

For the original problem, I want to flip the numerator and denominator in a way that I want get to the form of $\left(\dfrac{ab}{m}\right)$ Is there any way of doing this, or does this question require a completely different proof?

John W. Smith
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  • Legendre symbol is defined only when the denominator is prime. Do you mean the Jacobi Symbol? – N. S. Apr 24 '18 at 22:08
  • Yes, I got the two confused. Fixed. – John W. Smith Apr 24 '18 at 22:10
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    Isn't then what you want to prove the DEFINITION of the Jacobi symbol? – N. S. Apr 24 '18 at 22:12
  • Yes exactly. I was wondering if the proof would be analogous to the second case I gave, where $ab$ and $m$ interchanged. I showed the proof of the latter case for clarity purposes. – John W. Smith Apr 24 '18 at 22:15
  • @JohnSmith: I'm not sure what you have in mind with "analogous" -- I suspect you are trying to make the problem a lot harder than it actually is. The proof for the question you're asking about is actually fewer steps than the example proof. –  Apr 24 '18 at 22:19
  • Now that you mention it, you're probably right. How should I approach this question then? – John W. Smith Apr 24 '18 at 22:21
  • Wait. I figured it out. We simply by definition of the Jacobi symbol and we are done. – John W. Smith Apr 24 '18 at 22:25

1 Answers1

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As suggested in the comments, this is an easy consequence of the definition of the Jacobi symbol. But remember that the Jacobi symbol is defined as the product of Legendre symbols, so each 'denominator' needs to be prime.

So, let's suppose $a = p_1 \dotsm p_r$ and $b = q_1 \dotsm q_s$, for each $p_i,q_j$ prime. Then, \begin{align} \left(\dfrac{m}{a}\right) \left(\dfrac{m}{b}\right) &= \left(\dfrac{m}{p_1\dotsm p_r}\right) \left(\dfrac{m}{q_1\dotsm q_s}\right) \\ &=\left(\dfrac{m}{p_1}\right)\dotsm \left(\dfrac{m}{p_r}\right) \left(\dfrac{m}{q_1}\right) \dotsm \left(\dfrac{m}{q_s}\right) \\&=\left(\dfrac{m}{p_1 \dotsm p_r q_1\dotsm q_s}\right) \\ &= \left(\dfrac{m}{ab}\right) \end{align}

B. Mehta
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