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With $n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}$, both $a$ and $n$ as positive, odd integers, and defining the Jacobi symbol as (where on the right hand side we have the Legendre symbol):

$$\left( \frac{a}{n} \right) = \prod_{i=1}^k \left( \frac{a}{p_i} \right)^{e_i}$$

From this definition and the law of quadratic reciprocity for the Legendre symbol:

$$\left( \frac{a}{n} \right) = \prod_{i=1}^k \left( \frac{a}{p_i} \right)^{e_i} = \prod_{i=1}^k \left( (-1)^{\frac{a-1}{2} \cdot \frac{p_i-1}{2}} \left( \frac{p_i}{a} \right) \right)^{e_i}$$

$$\left( \frac{a}{n} \right) = \prod_{i=1}^k \left( (-1)^{\frac{a-1}{2} \cdot \frac{p_i - 1}{2}} \left( \frac{p_i}{a} \right) \right)^{e_i} = \left( \frac{n}{a} \right) \prod_{i=1}^k \left( (-1)^{\frac{a-1}{2} \cdot \frac{p_i - 1}{2}} \right)^{e_i} $$

So based on the law of quadratic reciprocity, we must now have:

$$(-1)^{\frac{a - 1}{2} \cdot \frac{n - 1}{2}} = \prod_{i=1}^k \left( (-1)^{\frac{a-1}{2} \cdot \frac{p_i - 1}{2}} \right)^{e_i} = \prod_{i=1}^k (-1)^{\frac{a-1}{2} \cdot \frac{p_i - 1}{2} \cdot e_i}$$

However, I don't see how this could be true, given that the right hand side is equivalent to saying that $\frac{n-1}{2} = \sum_{i=1}^k \frac{p_i - 1}{2} e_i$, when $n$ is also $p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}$. So I assume I did something wrong, but I'm not sure what that is.

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    $\begingroup$ You assumed $\left(\frac{a}{p_i}\right)=(-1)^{(a-1)(p_i-1)/4}\left(\frac{p_i}a\right)$. I can see that's true if $a$ is an odd prime, but not for general $a$. $\endgroup$ – Lord Shark the Unknown May 1 '17 at 14:38
  • $\begingroup$ @LordSharktheUnknown So if $a$ is an odd prime, then is $\frac{n - 1}{2} = \prod_{i=1}^k \frac{p_i - 1}{2} e_i$? If so, how? I don't see how that can be true. $\endgroup$ – Reed Oei May 1 '17 at 15:10
  • $\begingroup$ No, that equation never holds ... but you only need it (or something similar) modulo 2. $\endgroup$ – Lord Shark the Unknown May 1 '17 at 15:12

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