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How to calculate Jacobi Symbol $\left(\dfrac{27}{101}\right)$?

The book solution $\left(\dfrac{27}{101}\right) = \left(\dfrac{3}{101}\right)^3 = \left(\dfrac{101}{3}\right)^3 = (-1)^3 = -1$

My solution $\left(\dfrac{27}{101}\right) = \left(\dfrac{101}{27}\right) = \left(\dfrac{20}{27}\right) = \left(\dfrac{2^2}{27}\right) \cdot \left(\dfrac{5}{27}\right)$ $= (-1) \cdot \left(\dfrac{27}{5}\right) = (-1) \cdot \left(\dfrac{2}{5}\right) = (-1) \cdot (-1) = 1.$

Whenever I encounter $\left(\dfrac{2^b}{p}\right)$, I use the formula $(-1)^{\frac{p^2 - 1}{8}}$ I guess mine was wrong, but I couldn't figure out where? Any idea?

Thank you,

3 Answers 3

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$\big(\frac{4}{27}\big) = +1,$ not $-1$.

You can use the formula $\big( \frac {2^b}{m} \big) = (-1)^{(m^2-1)/8}$ only when $b$ is odd. When $b$ is even, $2^b$ is a square so the value is $+1$.

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    This is for clarification, but one could also interpret $\big(\frac{4}{27}\big)$ as $x^2\equiv 4\bmod 27$. Right? Then we know of such a x that fits the scenario, in this case x being two.2017-03-23
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Your mistake is calculating $\left(\dfrac{2^2}{27}\right)\:.\:$ But since you wish to use only $2$'s it's much simpler, viz.

$ \left(\dfrac{27}{101}\right)\ =\ \left(\dfrac{128}{101}\right)\ =\ \left(\dfrac{2}{101}\right)^{7}\ =\ -1 $

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I think it's better to make sure that the number in the lower case is a prime, since there are examples, if I remember rightly, that the Jacobi symbol is 1 but the corresponding quadratic congruence is not solvable; in addition, as already mentioned, you cannot say that $\left(\dfrac{2^b}{p}\right)\ = (-1)^{(p^2 -1)/8}$; it is a mistake without second thought, and I think it can be well avoided if you know the quadratic reciprocity law well, thanks.

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    Uh, I am pretty sure that I want a chance to improve my understanding, can someone explain the downvote? Thanks.2011-04-07