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Math Help - Integral coefficients

  1. #1
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    Integral coefficients

    Let f(x) be a polynomial with integral coefficients and a\equiv b \ \mbox{(mod m)}. Then f(a)\equiv f(b) \ \mbox{(mod m)}.

    It make since that if a, b are in the domain, then the co-domain will also be congruent of the same integers but how do I go about doing this?
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by dwsmith View Post
    Let f(x) be a polynomial with integral coefficients and a\equiv b \ \mbox{(mod m)}. Then f(a)\equiv f(b) \ \mbox{(mod m)}.

    It make since that if a, b are in the domain, then the co-domain will also be congruent of the same integers but how do I go about doing this?
    This is because given  a\equiv b\bmod{m} and  c\equiv d\bmod{m} then  aj\equiv bj\bmod{m},\; a^k\equiv b^k\bmod{m},\; a+c\equiv b+d\bmod{m} .

    Since a polynomial is a composition of those three operations, we're done.
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    I don't actually have to show anything to do with the polynomials? It is ok to say since
    Quote Originally Posted by chiph588@ View Post
    This is because given  a\equiv b\bmod{m} and  c\equiv d\bmod{m} then  aj\equiv bj\bmod{m},\; a^k\equiv b^k\bmod{m},\; a+c\equiv b+d\bmod{m} .

    Since a polynomial is a composition of those three operations, we're done.
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by dwsmith View Post
    I don't actually have to show anything to do with the polynomials? It is ok to say since
    I think it's enough work, but some might not.

    I'll walk you through an example:  f(x)=2x^2+3x+4

     a\equiv b\bmod{m}\implies a^2\equiv b^2\bmod{m} \implies 2a^2\equiv2b^2\bmod{m} (1)

     a\equiv b\bmod{m}\implies 3a\equiv3b\bmod{m} \implies 2a^2+3a\equiv2b^2+3b\bmod{m} from (1)

    and finally  2a^2+3a+4\equiv2b^2+3b+4\bmod{m} .

    If you wanted to get formal with this I suppose inducting on  \text{deg}(f) would do the trick.
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