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Math Help - Divisibility Proof

  1. #1
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    Divisibility Proof

    Show that if 7 | x^2 + 1, then 13 | x^3 + 5x^2 +17x - 100.

    I'm really not sure how to do this other than use the definition of divisibility, though I don't know where to go after that.
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  2. #2
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    Quote Originally Posted by Snooks02 View Post
    Show that if 7 | x^2 + 1, then 13 | x^3 + 5x^2 +17x - 100.

    I'm really not sure how to do this other than use the definition of divisibility, though I don't know where to go after that.
    The statement 7|(x^2+1) is never satisfied for any integer. Therefore, the conditional that 7|(x^2+1)\implies 13|(x^3 + 5x^2 + 17x^2 - 100) is immediately true.
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  3. #3
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    Quote Originally Posted by Snooks02 View Post
    Show that if 7 | x^2 + 1, then 13 | x^3 + 5x^2 +17x - 100.

    I'm really not sure how to do this other than use the definition of divisibility, though I don't know where to go after that.
    Have you checked to see if this true? In particular, what is the smallest value of x such that 7 divides x^2+ 1?
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  4. #4
    Super Member PaulRS's Avatar
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    If we had x^2\equiv{-1}(\bmod.7) (1) we see immediately that (x,7)=1 and then, by Fermat's little Theorem (*): x^6\equiv{1}(\bmod.7), however, (1) implies: x^6=(x^2)^3\equiv{(-1)^3}={-1}(\bmod.7) which is a contradiction!

    In fact there exists x\in \mathbb{Z} such that p|(x^2+1) ( p is a prime with p>2) if and only if p\equiv{1}(\bmod.4)
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