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.

  • 3
    $\begingroup$ 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
  • 2
    $\begingroup$ 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

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.


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.