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Math Help - need help on quadratic reciprocity

  1. #1
    Junior Member
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    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.
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  2. #2
    Senior Member
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    Atlanta, GA
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    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
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