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Math Help - Polynomial Congruences

  1. #1
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    Polynomial Congruences

    How do you find all the solutions of the congruence

    7x^2-41x+48=0(mod333)
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    Multiplying the congruence by 28 gives 196x^2-1148x+1344\equiv0\,(\bmod{}\,333).

    Note that \gcd(28,333)=1 and so the second congruence implies the first as well: i.e. 7x^2-41x+48\equiv0\,(\bmod{}\,333)\ \iff\ 196x^2-1148x+1344\equiv0\,(\bmod{}\,333).

    The reason for multiplying by 28 is to make the leading coefficient a perfect square and the coefficient of x even so that we can complete the square. Completing the square gives (14x-41)^2-337\equiv0\,(\bmod{}\,333). Hence, letting y=14x-41, we have to solve the congruence

    y^2\equiv337\,(\bmod{}\,333)\equiv4\,(\bmod{}\,333  )\quad\ldots\,\boxed1

    Clearly y=2 is a solution, so one solution to the original congruence is found by solving 14x-41\equiv2\,(\bmod{}\,333) for x. I can tell you that one possiblity is x=122 and substituting into the original congruence shows that x=122 is indeed one solution.

    Another solution to \boxed1 is y=-2. Find all solutions to \boxed1 and for each check and see if the corresponding solution for x satisfies the original congruence. Good luck.
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