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.
 A: 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.

A: 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. 
