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I have the following two Legendre symbols that need calculated:

$\left(\frac{10}{31}\right)$ $=$ $-\left(\frac{31}{10}\right)$ $=$ $-\left(\frac{1}{10}\right)$ $=$ $-(-1)$ $=$ $-1$

$\left(\frac{-15}{43}\right)$ $=$ $\left(\frac{43}{15}\right)$ $=$ $\left(\frac{13}{15}\right)$ $=$ $-1$

is that correct?

I just want to make sure I am understanding this concept.

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    $\begingroup$ If you are using the Legendre symbol (and not the Jacobi symbol), then the denominators are supposed to be prime. Clearly, $10$ and $15$ are not prime, and the quadratic reciprocity law you are using is highly dubious since it works only with odd integers. $\endgroup$ – Batominovski Dec 6 '18 at 22:35
  • $\begingroup$ @Batominovski ohh I did not know! Thank you so much for the comment $\endgroup$ – Hidaw Dec 6 '18 at 22:37
  • $\begingroup$ And please use a more descriptive title the next time. $\endgroup$ – Batominovski Dec 6 '18 at 22:37
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    $\begingroup$ Even with Jacobi symbols (as an intermediate step in the calculation) you need to treat factors of 2 on the top specially. $\endgroup$ – Daniel Schepler Dec 6 '18 at 22:45
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Here is an approach I would take. Note that $$\left(\frac{10}{31}\right)=\left(\frac{-21}{31}\right)=\left(\frac{-1}{31}\right)\left(\frac{3}{31}\right)\left(\frac{7}{31}\right)=(-1)\Biggl(-\left(\frac{31}{3}\right)\Biggr)\Biggl(-\left(\frac{31}{7}\right)\Biggr)\,.$$

That is, $$\left(\frac{10}{31}\right)=-\left(\frac{1}{3}\right)\left(\frac{3}{7}\right)=-(+1)\Biggl(-\left(\frac{7}{3}\right)\Biggr)=\left(\frac{1}{3}\right)=+1\,.$$ This can be verified by noting that $6^2\equiv 5\pmod{31}$ and $8^2\equiv 2\pmod{31}$, so $$17^2\equiv (6\cdot 8)^2\equiv 5\cdot2=10\pmod{31}\,.$$


For the second part, note that $$\left(\frac{-15}{43}\right)=\left(\frac{-1}{43}\right)\left(\frac{3}{43}\right)\left(\frac{5}{43}\right)=(-1)\Biggl(-\left(\frac{43}{3}\right)\Biggr)\left(\frac{43}{5}\right)\,.$$

Therefore, $$\left(\frac{-15}{43}\right)=\left(\frac{1}{3}\right)\left(\frac{3}{5}\right)=\left(\frac{3}{5}\right)\,.$$ It is easy to verify that $\left(\dfrac{3}{5}\right)=-1$, whence $$\left(\frac{-15}{43}\right)=-1\,.$$ You can check that $12^2\equiv 15\pmod{43}$, so $\left(\dfrac{15}{43}\right)=+1$, whereas $\left(\dfrac{-1}{43}\right)=-1$, confirming the calculations.

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As remarked in comments, to use quadratic reciprocity we need to work with Legendre Symbols $$ \left(\frac{p}{q}\right)$$ for $p,q$ prime. You should repeatedly use the property $$ \left(\frac{ab}{p}\right)=\left(\frac{a}{p}\right)\left(\frac{b}{p}\right)$$ to make sure that you are calculating with both parts of the symbol being prime. That is, write $$ \left(\frac{10}{31}\right)=\left(\frac{2}{31}\right)\left(\frac{5}{31}\right)$$ then iteratively apply quadratic reciprocity as you intended to.

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  • $\begingroup$ But not to $2$ if I recall correctly. You can render $(2|31)=1$ from $8^2=64$, but that is not QR. $\endgroup$ – Oscar Lanzi Dec 6 '18 at 23:23
  • $\begingroup$ If you don't mind could you please expand on that. I am trying solve it but I missing something and I cannot figure out what $\endgroup$ – Hidaw Dec 7 '18 at 20:28

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