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Math Help - Solve the Quadratic Congruence

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    Solve the Quadratic Congruence

    3x2 + 2x ≡ 4 mod 17
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  2. #2
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    Quote Originally Posted by mndi1105 View Post

    3x^2 + 2x \equiv 4 \mod 17
    since 6 and 17 are coprime, the above equation is equivalent to: 18x^2 + 12 x \equiv 24 \mod 17, which is equivalent to: x^2 + 12 x - 7 \equiv 0 \mod 17.

    now we have: x^2 + 12x - 7=(x+6)^2 - 43 \equiv (x+6)^2 - 9 \mod 17. but (x+6)^2 - 9 = (x+3)(x+9). so the solutions to your equation are:

    x \equiv -3 \equiv 14 \mod 17 and x \equiv -9 \equiv 8 \mod 17. \ \ \ \Box
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