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Polynomials (mod p)

  1. Dec 15, 2004 #1
    I'm having problems finding all integer solutions to some of the higher degree polynomials.

    for p(x)= x^3− 3x^2+ 27 ≡ 0 (mod 1125), i get that 1125 = (3^2)(5^3).
    p(x) ≡ 0 (mod 3^2), p(x) ≡ 0 (mod 5^3).
    x ≡ 0, 3, 6 (mod 3^2) for 3^2
    for 5^3, x ≡ 51 (mod 5^3)
    then i get x=801, 51, 426 (mod 1125).

    but i cannot seem to get as eloquent of an answer for p(x)= 4x^4 + 9x^3 - 5x^2 - 21x + 61.

    can anyone help? i know you start out the same way. perhaps there is an easier way?
     
  2. jcsd
  3. Dec 16, 2004 #2

    shmoe

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    What snag are you running into? Everything should work out the same way. Were you able to find zeros mod 5^3 and mod 3^2? (I'm assuming the same modulus for both questions)
     
  4. Dec 16, 2004 #3
    maybe i don't have the right zeros... i'll try again and see what i get.
     
  5. Dec 22, 2004 #4
    for the zeros, i got:

    b=3, 1, 2

    i got no solutions for b= 1,2
    for b= 3 i got x=8 (mod 5)
     
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