Solving Polynomials (mod p) Problems

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In summary, the conversation is about finding all integer solutions for higher degree polynomials, specifically p(x)= x^3− 3x^2+ 27 ≡ 0 (mod 1125). The summary outlines the process of finding solutions using modular arithmetic and states the final solutions as x=801, 51, 426 (mod 1125). The conversation then shifts to a different polynomial, p(x)= 4x^4 + 9x^3 - 5x^2 - 21x + 61, and the speaker asks for help in finding the solutions. After discussing the approach, the speaker mentions getting b=3, 1, 2 as potential zeros but no solutions for b
  • #1
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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  • #2
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)
 
  • #3
maybe i don't have the right zeros... i'll try again and see what i get.
 
  • #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)
 

FAQ: Solving Polynomials (mod p) Problems

What is the meaning of "mod p" in solving polynomials problems?

"Mod p" refers to the process of performing arithmetic operations on integers but only considering the remainder when divided by a prime number, p. This is also known as modular arithmetic.

How does solving polynomials (mod p) differ from traditional polynomial solving?

In traditional polynomial solving, we are working with real numbers and the operations are performed using the rules of algebra. However, in solving polynomials (mod p) problems, we are working with integers and the operations are performed using the rules of modular arithmetic.

What is the significance of using modular arithmetic in solving polynomials (mod p) problems?

Modular arithmetic allows us to work with large numbers more efficiently and effectively. It also has applications in cryptography and computer science.

Can you provide an example of solving a polynomial (mod p) problem?

Sure, let's say we have the polynomial 3x^3 + 2x^2 + 5x + 1 and we want to find its value when x = 2 (mod 7). We would first substitute 2 for x and perform the operations in modulo 7. So, (3*2^3 + 2*2^2 + 5*2 + 1) mod 7 = (24 + 8 + 10 + 1) mod 7 = 5 mod 7 = 5. Therefore, the value of the polynomial when x = 2 (mod 7) is 5.

How do I know when to use solving polynomials (mod p) versus traditional polynomial solving?

Solving polynomials (mod p) is typically used in cases where we are working with integers or large numbers and need to perform operations efficiently. It is also useful in certain fields such as number theory, cryptography, and computer science. Traditional polynomial solving is used in most other cases involving real numbers and algebraic operations.

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