What are the Legendre symbols $\left(\frac{10}{31}\right)$ and $\left(\frac{-15}{43}\right)$?

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.

• 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. – Batominovski Dec 6 '18 at 22:35
• @Batominovski ohh I did not know! Thank you so much for the comment – Hidaw Dec 6 '18 at 22:37
• And please use a more descriptive title the next time. – Batominovski Dec 6 '18 at 22:37
• Even with Jacobi symbols (as an intermediate step in the calculation) you need to treat factors of 2 on the top specially. – 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.

• But not to $2$ if I recall correctly. You can render $(2|31)=1$ from $8^2=64$, but that is not QR. – Oscar Lanzi Dec 6 '18 at 23:23
• 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 – Hidaw Dec 7 '18 at 20:28