# Verification: have I worked this legendre symbol correctly

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?

• $92 \equiv 4 \bmod 11$ – reuns Aug 3 '16 at 11:57
• Why are you looking at quadratic residue here anyway? – Bill Wallis Aug 3 '16 at 12:17
• @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 – reuns Aug 3 '16 at 12:55

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