# need help on quadratic reciprocity

• Apr 8th 2009, 04:03 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
\$\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.
• Apr 15th 2009, 01:01 PM
Media_Man
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