Solving Polynomials (mod p) Problems

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SUMMARY

The discussion focuses on solving polynomial equations modulo a composite number, specifically addressing the polynomial p(x) = x^3 - 3x^2 + 27 ≡ 0 (mod 1125) and the challenges faced with p(x) = 4x^4 + 9x^3 - 5x^2 - 21x + 61. The user successfully identifies integer solutions for the first polynomial using the Chinese Remainder Theorem, finding solutions x ≡ 801, 51, 426 (mod 1125). However, they encounter difficulties in determining solutions for the second polynomial, indicating a need for further exploration of roots modulo 3^2 and 5^3.

PREREQUISITES
  • Understanding of modular arithmetic, particularly with composite moduli.
  • Familiarity with the Chinese Remainder Theorem.
  • Basic knowledge of polynomial equations and their roots.
  • Experience with integer solutions in modular systems.
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  • Research the application of the Chinese Remainder Theorem for higher degree polynomials.
  • Learn techniques for finding roots of polynomials modulo prime powers, specifically mod 3^2 and mod 5^3.
  • Explore numerical methods for solving polynomial equations in modular arithmetic.
  • Investigate software tools like SageMath for computational assistance in polynomial root finding.
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Mathematicians, computer scientists, and students engaged in number theory, particularly those focused on polynomial equations and modular arithmetic.

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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