-1
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Compute $(\frac{92}{11}$).

Now $92^2 \equiv x^2\ mod(11) $ Now since there is no such x that satisfies this, so the legendre symbol is $-1$. Is this right?

Also, can somebody explain in a slightly more layman's language what the significance of quadratic residues are?

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  • $\begingroup$ $92 \equiv 4 \bmod 11$ $\endgroup$ – reuns Aug 3 '16 at 11:57
  • $\begingroup$ Why are you looking at quadratic residue here anyway? $\endgroup$ – Bill Wallis Aug 3 '16 at 12:17
  • $\begingroup$ @BillWallis ?? $(a \mid p) \equiv a^{(p-1)/2} \bmod p$ but it is also $1$ if $a \equiv x^2$, $-1$ otherwise Legendre symbol therefore $92 \equiv 4 \equiv 2^2 \bmod 11$ is enough for answering $\endgroup$ – reuns Aug 3 '16 at 12:55
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You can use the Legendre formula $\left( a \over p \right)=a^{(p-1)/2} \pmod p $ to manually compute the value. $\bmod 11$ you have: $$\left( 92 \over 11 \right)\equiv 92^5 \equiv 4^5 \equiv 4\times 16^2 \equiv 4\times 25 \equiv 100 \equiv 1$$ Therefor you value $-1$ is wrong.

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  • $\begingroup$ How did you get this step? $4×16^2$≡4×25 $\endgroup$ – stackdsewew Aug 3 '16 at 15:04
  • $\begingroup$ $16\equiv 5 \pmod {11}$ $\endgroup$ – gammatester Aug 4 '16 at 6:30

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