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Math Help - Legendre Symbol

  1. #1
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    Legendre Symbol

    This question has completely floored me.
    __________________________________

    Let f(x) = x^7 -2x^6-2x^5-3x^4. Evaluate the Legendre symbol

    \displaystyle \left( \frac{f(x)}{19} \right)

    for all integers x.
    __________________________________

    I have factorised f(x) and have the following, by multiplicativity:

    \displaystyle \left( \frac{f(x)}{19} \right) = \left( \frac{x}{19} \right)^4 \left( \frac{x-3}{19} \right) \left( \frac{x^2 + x +1}{19} \right)

    and the first term must be +1 (as it is raised to an even power). But I've been manipulating the other two for the last two hours and have gotten pretty much nowhere. Any ideas?
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  2. #2
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    Quote Originally Posted by Capillarian View Post
    This question has completely floored me.
    __________________________________

    Let f(x) = x^7 -2x^6-2x^5-3x^4. Evaluate the Legendre symbol

    \displaystyle \left( \frac{f(x)}{19} \right)

    for all integers x.
    __________________________________

    I have factorised f(x) and have the following, by multiplicativity:

    \displaystyle \left( \frac{f(x)}{19} \right) = \left( \frac{x}{19} \right)^4 \left( \frac{x-3}{19} \right) \left( \frac{x^2 + x +1}{19} \right)

    and the first term must be +1 (as it is raised to an even power). But I've been manipulating the other two for the last two hours and have gotten pretty much nowhere. Any ideas?

    x^2+x+1=(x+8)(x+12)\!\!\pmod{19} , so taking into account your first deduction we get

    \left(\frac{f(x)}{19}\right)=\left(\frac{x+16}{19}  \right)\left(\frac{x+8}{19}\right)\left(\frac{x+12  }{19}\right).

    Now input all the values from 0\,\,to\,\,18\!\!\pmod{19} , and with a list of the squares modulo 19 by your side check each output...

    Tonio
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  3. #3
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    Edit: Terrible formatting. See below.
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  4. #4
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    Wonderful! Thanks so much. That question was bugging me for ages. I got the following:

    \displaystyle \left( \frac{f(x)}{19} \right) = \begin{Bmatrix} 0 & \text{ if } x \equiv 0, 4, 7, \text{ or } 11 (\mod 19) \\ +1 & \text{ if } x \equiv 2, 3, 8, 10, 14, \text{ or } 17 (\mod 19) \\ -1 & \text{ otherwise. } \end{matrix}

    Does that seem the right sort of thing?
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  5. #5
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    Quote Originally Posted by Capillarian View Post
    Wonderful! Thanks so much. That question was bugging me for ages. I got the following:

    \displaystyle \left( \frac{f(x)}{19} \right) = \begin{Bmatrix} 0 & \text{ if } x \equiv 0, 4, 7, \text{ or } 11 (\mod 19) \\ +1 & \text{ if } x \equiv 2, 3, 8, 10, 14, \text{ or } 17 (\mod 19) \\ -1 & \text{ otherwise. } \end{matrix}

    Does that seem the right sort of thing?

    Why do you get zero with 4\!\!\pmod {19} ? I get 1...Likewise, with 2\!\!\pmod {19} I get -1 ...Check all these.

    Tonio
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  6. #6
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    Quote Originally Posted by tonio View Post
    Why do you get zero with 4\!\!\pmod {19} ? I get 1...Likewise, with 2\!\!\pmod {19} I get -1 ...Check all these.

    Tonio
    Argh! I've accidentally written the factorization down as f(x) = (x+8)(x+12)(x+15) \mod 19 when the last term should have been 16. That's it -- I get the same answer as you now. Thanks again!
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