Legendre symbol of $\left(\!\frac{11}{71}\!\right)$ I'm trying to find the Legendre symbol of 
$\left(\!\frac{11}{71}\!\right)$ . 
Here is what I did so far : 
$$\left(\!\frac{11}{71}\!\right)=(-1)^{{((11-1)/2}⋅{(71-1)/2)}}⋅\left(\!\frac{71}{11}\!\right)=(-1)^{5} (-1)^{35} \left(\!\frac{71}{11}\!\right)$$
$$\left(\!\frac{11}{71}\!\right)= 71^{((11-1)/2)} \pmod{11}=71^5 \pmod{11} = ?$$
How can I continue from here ? 
Thanks
 A: You use the law of quadratic reciprocity which says that 
$$\left(\frac{p}{q}\right)=\left(\frac{q}{p}\right)$$ except when both $p$ and $q$ are $\equiv 3\pmod 4$ in which case 
$$\left(\frac{p}{q}\right)=-\left(\frac{q}{p}\right)$$
So 
$$\left(\frac{11}{71}\right)=-\left(\frac{71}{11}\right)=-\left(\frac{5}{11}\right)=
-\left(\frac{11}{5}\right)=-\left(\frac{1}{5}\right)=-1$$
Where we note that $$11,71 \equiv 3 \pmod 4$$ and $$5 \equiv 1 \pmod 4$$
A: Although $71^5 = 1804229351$ is technically a small number by the frivolous theorem of arithmetic, it can still be daunting, especially if you don't have a computer or at least a calculator.
But what you can do is go through the exponents modulo $11$ from the get-go. Thus:


*

*$71 \equiv 5 \pmod{11}$

*$71^2 \equiv 5^2 \equiv 3 \pmod{11}$

*$71^3 \equiv 3 \times 5 \equiv 4 \pmod{11}$

*$71^4 \equiv 4 \times 5 \equiv 9 \pmod{11}$

*$71^5 \equiv 9 \times 5 \equiv 1 \pmod{11}$


Therefore $$\left(\frac{11}{71}\right) = 1.$$
P.S. In case of the opposite answer, depending on how you went about it, you might see $10$ rather than $-1$. It still means the same thing.
A: For the first attempt, as noted in the comments, you have an error: $(-1)^{ab}\ne(-1)^a(-1)^b$.
For both attempts, also as noted in the comments, to continue further reduce $71$ mod $11$.
Here is a different example to get an idea with:
$$\left(\frac{13}{59}\right)=(-1)^{6\cdot29}\left(\frac{59}{13}\right)=\left(\frac{7}{13}\right)=(-1)^{3\cdot6}\left(\frac{13}{7}\right)=\left(\frac{-1}{7}\right)=(-1)^{3}. $$
