Legendre symbol simplification I saw a simplification using the Legendre symbol which said $$\left(\dfrac{19}{29}\right) = \left(\dfrac{10}{19}\right) = \left(\dfrac{2}{19}\right) \cdot \left(\dfrac{5}{19}\right) = -1.$$ My question is how did they get $\left(\dfrac{19}{29}\right) = \left(\dfrac{10}{19}\right)$ and finally that $\left(\dfrac{2}{19}\right) \cdot \left(\dfrac{5}{19}\right) = -1$?
 A: For the first question
$$
\left(\dfrac{19}{29}\right) = \left(\dfrac{29}{19}\right)
 = \left(\dfrac{10}{19}\right)
$$
since $29$ is $1$ mod $4$.
For the second, 
$$
\left(\dfrac{5}{19}\right) = 
\left(\dfrac{19}{5}\right) = 
\left(\dfrac{-1}{5}\right) = 1
$$
since $5$ is $1$ mod $4$
and
$$
\left(\dfrac{2}{19}\right) = -1
$$
because $19$ is $3$ mod $8$. (http://www.maa.org/programs/faculty-and-departments/classroom-capsules-and-notes/the-quadratic-character-of-2)
A: This seems a combination of quadratic reciprocity and the Legendre formula.
The Legendre formula states that $\left(\dfrac{a}{p}\right) \equiv a^{\frac{p-1}{2}}\mod p$, giving $\left(\dfrac{2}{19}\right) = -1$ and $\left(\dfrac{5}{19}\right) = 1$. These give the second equality. (Since 19 isn't that large, it's quite possible to do determine by hand if 2 or 5 are quadratic residues).
The quadratic reciprocity law states that $\left(\dfrac{p}{q}\right)\left(\dfrac{q}{p}\right) = (-1)^{{\frac{p-1}{2}}{\frac{q-1}{2}}}$. Since $\left({\dfrac{29}{19}}\right) = \left({\dfrac{10}{19}}\right)$, the first equality you asked for follows. (Notice that since $\left(\dfrac{p}{q}\right)$ and its reciprocal are $\pm1$, dividing or multiplying by one of them yields the same result.)
