Polynomials (mod p)

  • Thread starter ascheras
  • Start date
  • #1
14
0
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?
 

Answers and Replies

  • #2
shmoe
Science Advisor
Homework Helper
1,992
1
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
14
0
maybe i don't have the right zeros... i'll try again and see what i get.
 
  • #4
6
0
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)
 

Related Threads on Polynomials (mod p)

  • Last Post
Replies
4
Views
7K
  • Last Post
Replies
9
Views
4K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
7
Views
8K
  • Last Post
Replies
4
Views
6K
  • Last Post
Replies
3
Views
4K
Replies
5
Views
6K
  • Last Post
Replies
12
Views
4K
Replies
2
Views
7K
Top