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Math Help - Congruence, prime number, integer

  1. #1
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    Congruence, prime number, integer

    let p be odd prime, and a,b be integers
    1) prove that the congruence {x}^{2}-ab\equiv (a-b)x(mod p) always have one solution
    2) when does the above congruence have 2 solutions modulo p?
    3) solve the congruence 4{x}^{3}\equiv x(mod p)
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  2. #2
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    Quote Originally Posted by Shurelia View Post
    let p be odd prime, and a,b be integers
    1) prove that the congruence {x}^{2}-ab\equiv (a-b)x(mod p) always have one solution
    2) when does the above congruence have 2 solutions modulo p?
    3) solve the congruence 4{x}^{3}\equiv x(mod p)
    The wording for part one is strange to me.

    Consider

    x^2-ab \equiv (a -b)x \mod{p} \iff x^2 +(b-a)x-ab \equiv 0 \mod{p}

    Since the integers mod p are a finite field they are an integral domain we have that

     (x+b)(x-a) \equiv 0 \mod{p}

    I cannot see any reason why [tex]a \equiv -b \mod{p}[\tex] So it looks like it can have two solutions.

    Please clarify and see if this gets you started on the next problem.

    For the last one just factor!
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