# Polynomials (mod p)

1. Dec 15, 2004

### ascheras

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. Dec 16, 2004

### shmoe

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. Dec 16, 2004

### ascheras

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

4. Dec 22, 2004

### brute26

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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