Evaluate the Legendre symbols $(\frac{503}{773})$ and $(\frac{501}{773})$ Evaluate the Legendre symbols $(\frac{503}{773})$ and $(\frac{501}{773})$.
My solution:
$(\frac{501}{773}) = (\frac{167 \cdot 3}{773}) = (\frac{167}{773}) \cdot (\frac{3}{773}) = (\frac{773}{167}) \cdot (\frac{773}{3})$
$= (\frac{105}{167}) \cdot (\frac{2}{3})$
$= (\frac{3}{167}) \cdot (\frac{5}{167}) \cdot (\frac{7}{167}) \cdot (\frac{2}{3})$ 
$=-1 \cdot (\frac{167}{3}) \cdot (\frac{167}{5}) \cdot -1 \cdot (\frac{167}{7}) \cdot (\frac{2}{3})$
$= (\frac{2}{3}) \cdot (\frac{2}{5}) \cdot (\frac{6}{7}) \cdot (\frac{2}{3})$ 
$= (\frac{2}{3}) \cdot (\frac{2}{5}) \cdot (\frac{2}{7}) \cdot (\frac{3}{7}) \cdot (\frac{2}{3})$ 
$= -1 \cdot -1 \cdot 1 \cdot -1 \cdot -1$
$=1$ 
Is it correct? I need confirmation if it is correct or not, and help in the other part.

The calculation for $501$ is correct. 
You should find that $503$ is a quadratic non-residue of $773$. The calculation is easier, because $503$ is prime and $270$ has lots of small factors. We have
$$(503/773)=(773/503)=(270/503)=(2/503)(3/503)(5/503).$$
Because $503$ is of the form $8k+7$, we have $(2/503)=1$. Easily, $(3/503)=-(503/3)=1$. And finally $(5/503)=(503/5)=(3/5)=-1$.
