<s>[INST] how did we find the Legendre symbol? The Legendre symbol from a prime $p>2$ and $(a,p)=1$ is:
$$\left ( \frac{a}{p} \right )=\left\{\begin{matrix}
1, & \text{if a is a quadratic residue} \\ 
 -1& \text{if a is a non-quadratic residue}
\end{matrix}\right.$$
According to my notes for $p=11$:
$$\left ( \frac{1}{11} \right )=\left ( \frac{3}{11} \right )=\left ( \frac{4}{11} \right )=\left ( \frac{5}{11} \right )=\left ( \frac{9}{11} \right )=1$$
$$\left ( \frac{2}{11} \right )=\left ( \frac{6}{11} \right )=\left ( \frac{7}{11} \right )=\left ( \frac{8}{11} \right )=\left ( \frac{10}{11} \right )=-1$$
How did we find the Legendre symbol?
 [/INST] What Dietrich said! :)
Alternatively, there are a number of ways to evaluate a Legendre symbol.
For starters you can apply its definition.
Since $x^2 \equiv 1 \pmod {11}$ has the solution $x\equiv 1 \pmod {11}$, it follows that:
$$\left({1\over 11}\right) = 1$$
Or alternatively you can apply the property:
$$\left({a\over p}\right) \equiv a^{\frac{p-1}2} \pmod p$$
We can apply it and find for instance:
$$\left({3\over 11}\right) \equiv 3^{\frac{11-1}2} \equiv 3^5 \equiv 1 \pmod{11}$$
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