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Math Help - Congruence equation

  1. #1
    Senior Member I-Think's Avatar
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    Congruence equation

    Good day to the MHF. Here's my problem
    Find all integer solutions to x^{19}+3x^2\equiv10 (mod 21)

    Just making sure my solution is complete
    Solution
    x^{19}+3x^2\equiv10 (mod 21)
    By CRT, we can break this up to
    x^{19}+3x^2\equiv10 (mod 3)
    x^{19}+3x^2\equiv10 (mod 7)

    And now we deal with each separately
    For x^{19}+3x^2\equiv10 (mod 3)
    we use FLT to simplify: x^2\equiv 1 (mod 3)
    Simplifying
    x+3\equiv10 (mod 3). Solves to give x\equiv7 (mod 3)

    Now dealing with
    x^{19}+3x^2\equiv10 (mod 7)
    FLT: x^6\equiv1 (mod 7)
    Simplifying
    3x^2+x\equiv10 (mod 7) \rightarrow, we solve this via guesswork

    Let 3x^2+x-10=14\rightarrow3x^2+x-24=0
    (3x-8)(x+3)=0<br />
Solution: x\equiv-3 (mod 7)

    So now we use CRT to solve
    x\equiv7 (mod 3)
    x\equiv-3 (mod 7)
    x-7=3m<br />
x+3=7n

    7n-3m=10 Solve to get m=20, n=10
    So x=67

    And a solution to our original equation is  x\equiv67 (mod 21)

    Is this all the solutions or am I missing some? If so, where is te error in my method preventing me from obtaining the others?
    Thanks in advance for the help.
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  2. #2
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    Hello, I-Think!

    You're off to a good start . . .


    Find all integer solutions to: . x^{19}+3x^2\:\equiv\:10\text{ (mod 21)}


    Just making sure my solution is complete

    Solution

    x^{19}+3x^2\equiv10 (mod 21)

    By CRT, we can break this up to: . \begin{Bmatrix}x^{19}+3x^2\:\equiv\:10 \text{ (mod 3)} \\<br />
x^{19}+3x^2\:\equiv\:10 \text{ (mod 7)} \end{Bmatrix}


    And now we deal with each separately

    For x^{19}+3x^2 \:\equiv\:10 \text{ (mod 3)}
    we use FLT to simplify: . x^2\:\equiv\: 1 \text{ (mod 3)}

    Simplifying: . x+3 \:\equiv\:10 \text{ (mod 3)}.

    Solve to give: . x\:\equiv\:7 \text{ (mod 3)}

    Hence, we have: . \boxed{x \:\equiv\:1\text{ (mod 3)}}




    Now dealing with: . x^{19}+3x^2\:\equiv\:10 \text{ (mod 7)}

    FLT: . x^6\:\equiv\:1 \text{ (mod 7)}

    Simplifying: . 3x^2+x \:\equiv\:10 \text{ (mod 7)}

    We have: . 3x^2 + x - 10 \:\equiv\:0\text{ (mod 7)}

    Factor: . (x+2)(3x-5) \:\equiv\:0\text{ (mod 7)}


    . . x+2\:\equiv\:0\text{ (mod 7)} \quad\Rightarrow\quad x \:\equiv\:-2\text{ (mod 7)} \quad\Rightarrow\quad \boxed{x \:\equiv\:5\text{ (mod 7)}}


    . . 3x-5\:\equiv\:0\text{ (mod 7)} \quad\Rightarrow\quad 3x \:\equiv\:5\text{ (mod 7)}

    . . Multiply by 5: . 15x\:\equiv\:25\text{ (mod 7)} \quad\Rightarrow\quad \boxed{x \:\equiv\:4\text{ (mod 7)}}

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