Solving Polynomials (mod p) Problems

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    Polynomials
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Discussion Overview

The discussion revolves around finding integer solutions to higher degree polynomials under modular arithmetic, specifically focusing on polynomials modulo 1125. Participants explore methods for solving these equations and share their experiences with specific polynomial examples.

Discussion Character

  • Exploratory, Technical explanation, Homework-related

Main Points Raised

  • One participant presents the polynomial p(x) = x^3 - 3x^2 + 27 and describes their process for finding solutions modulo 1125, breaking it down into components modulo 3^2 and 5^3.
  • They report finding solutions x ≡ 0, 3, 6 (mod 3^2) and x ≡ 51 (mod 5^3), leading to combined solutions x = 801, 51, 426 (mod 1125).
  • The same participant expresses difficulty in finding solutions for another polynomial, p(x) = 4x^4 + 9x^3 - 5x^2 - 21x + 61, and seeks assistance.
  • Another participant questions the first participant about the specific issues they are encountering and encourages them to verify their zeros for both moduli.
  • A later reply indicates that the first participant may not have the correct zeros and suggests they will attempt to find them again.
  • Subsequently, the first participant shares their findings for zeros, noting b = 3, 1, 2, and mentions having no solutions for b = 1, 2, while finding x = 8 (mod 5) for b = 3.

Areas of Agreement / Disagreement

The discussion reflects uncertainty and ongoing exploration, with no consensus reached on the methods or solutions for the second polynomial. Participants are actively sharing their findings and troubleshooting their approaches.

Contextual Notes

Participants have not fully resolved the methods for finding zeros for the second polynomial, and there may be missing assumptions regarding the approach to modular arithmetic in this context.

ascheras
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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?
 
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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)
 
maybe i don't have the right zeros... i'll try again and see what i get.
 
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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