need help on quadratic reciprocity

**mancillaj3** · April 8th 2009, 04:03 PM

Prove that the quadratic residues of 11 are 1,3,4,5,9, and list all solutions of each of the ten congruences
$x^{2} \cong a(\bmod 11)$ and $x^{2}\cong a(\bmod 11^{2})$ where a=1,3,4,5,9.

i used Legendre to show (1/11)=1,...,(9/11)=1, for $11^2$ i used legendre again to show a=1,3,4,5,9 are quadratic residues of $11^2$ .
My problem is with the second part have no idea how to even begin.

**Media_Man** · April 15th 2009, 01:01 PM

$0^2 \cong 0$ , $1^2 \cong 1$ , $2^2 \cong 4$ , $3^2 \cong 9$ , $4^2 \cong 5$ , $5^2 \cong 3$ , $6^2 \cong 3$ , $7^2 \cong 5$ , $8^2 \cong 9$ , $9^2 \cong 4$ , $10^2 \cong 1(\bmod 11)$

And $(x+11)^2=x^2+22x+121\cong x^2(\bmod 11)$

So the solutions for the first part are:
$x^{2} \cong 1(\bmod 11)$ x=1+11k or x=10+11k
$x^{2} \cong 3(\bmod 11)$ x=5+11k or x=6+11k
$x^{2} \cong 4(\bmod 11)$ x=2+11k or x=9+11k
$x^{2} \cong 5(\bmod 11)$ x=4+11k or x=7+11k
$x^{2} \cong 9(\bmod 11)$ x=3+11k or x=8+11k

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need help on quadratic reciprocity

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