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Not sure what's going wrong with the calculation here, trying to calculate the Legendre Symbol

$\left(\frac{138}{461}\right)$.

My process follows as $\left(\frac{138}{461}\right) = \left(\frac{2}{461}\right)\left(\frac{3}{461}\right)\left(\frac{23}{461}\right) = \left(\frac{461}{2}\right)\left(\frac{461}{3}\right)\left(\frac{461}{23}\right) = \left(\frac{1}{2}\right)\left(\frac{2}{3}\right)\left(\frac{1}{23}\right) = 1*-1 *1 = -1$

But $x^2 \equiv 138\ mod\ 461$ is soluble so something must have went wrong, not sure exactly what.

First I use the fact that if $a \equiv a' (mod\ p)$, then $\left(\frac{a}{p}\right) = \left(\frac{a'}{p}\right)$.

Then I use quadratic reciprocity since $461 \equiv 1\ mod\ 4$. From then on, I simply evaluate it (ex. 461 mod 2) and get $-1$. But the answer should be $1$ so I'm not sure where I went wrong.

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You should do the prime 2 separately.

$$ 461 \equiv 5 \pmod 8 $$ $$ (2|461) = -1 $$

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  • $\begingroup$ Mind elaborating based on the steps I wrote out? $\endgroup$ – SS' Mar 22 '18 at 3:44
  • $\begingroup$ @SS' there is no Legendre symbol $(a|p)$ when $p=2.$ $\endgroup$ – Will Jagy Mar 22 '18 at 3:47
  • $\begingroup$ Oh yeah, silly mistake. Thanks. $\endgroup$ – SS' Mar 22 '18 at 3:48

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