1. ## 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
$\displaystyle x^{2} \cong a(\bmod 11)$ and $\displaystyle x^{2}\cong a(\bmod 11^{2})$ where a=1,3,4,5,9.

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

2. ## Quadratic residues form a repeating pattern

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

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

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