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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?

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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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    $\begingroup$ I understand..thank you very much!!!! $\endgroup$ – evinda May 13 '14 at 20:53
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Compute all the squares $1^2,2^2,\ldots ,10^2$ modulo $11$. Then you will see that $1,3,4,5,9$ are quadratic residues, i.e., appearing in your list, and the other ones not. For example, $4^2\equiv 5$ mod $11$, hence $5$ is a quadratic residue, i.e., $(5/11)=1$.

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  • $\begingroup$ I understand...thanks a lot!!!! $\endgroup$ – evinda May 13 '14 at 20:52

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