# need help on quadratic reciprocity

• Apr 8th 2009, 05:03 PM
mancillaj3
need help on quadratic reciprocity
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.
• Apr 15th 2009, 02:01 PM
Media_Man
Quadratic residues form a repeating pattern
$0^2 \cong 0$, $1^2 \cong 1$, $2^2 \cong 4$, $3^2 \cong 9$, $4^2 \cong 5$, $5^2 \cong 3$, $6^2 \cong 3$, $7^2 \cong 5$, $8^2 \cong 9$, $9^2 \cong 4$, $10^2 \cong 1(\bmod 11)$

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