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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,

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4 Answers 4

up vote 3 down vote accepted

$\big(\frac{4}{27}\big) = +1,$ 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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Thanks, Is it $\frac{27^2-1}{8} = 91$? –  Chan Apr 5 '11 at 23:30

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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That formula works for $(\frac{2}{p})$, but $(\frac{4}{p})=1$, always.

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I calculated $(-1)^{\frac{27^2 - 1}{8}}$. Thanks, I got it ;)! –  Chan Apr 5 '11 at 23:32

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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1  
Uh, I am pretty sure that I want a chance to improve my understanding, can someone explain the downvote? Thanks. –  awllower Apr 7 '11 at 13:14

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