Polynomial divisor in $mathbb{Z}_p[x]$?
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Q: Find an odd prime $p$ for which $x-2$ is a divisor of $x^4 + x^3 + 3x^2 + x + 1$ in $mathbb{Z}_p[x]$.
I would rather not go through the Euclidean Algorithm for every mod $p$. Is there another way to find this particular prime?
Any help would be greatly appreciated.
abstract-algebra polynomials prime-numbers factoring
edited Dec 3 at 4:00
André 3000
12.2k22041
asked Dec 3 at 2:11
Raul Quintanilla Jr.
552
add a comment |
up vote
2
down vote
favorite
Q: Find an odd prime $p$ for which $x-2$ is a divisor of $x^4 + x^3 + 3x^2 + x + 1$ in $mathbb{Z}_p[x]$.
I would rather not go through the Euclidean Algorithm for every mod $p$. Is there another way to find this particular prime?
Any help would be greatly appreciated.
abstract-algebra polynomials prime-numbers factoring
edited Dec 3 at 4:00
André 3000
12.2k22041
asked Dec 3 at 2:11
Raul Quintanilla Jr.
552
8
note that $x-2$ divides $f(x)$ specifically when $f(2) = 0$, so plug in $x=2$ and find a prime divisor of the result.
– Rolf Hoyer
Dec 3 at 2:14
Rolf Hoyer has given a nice trick, but even doing this problem the "straightforward way" is quite short. We perform one polynomial division with remainder in $mathbb{Z}[x]$. The remainder will be a constant since the divisor has degree $1$, and then we pick a prime dividing the remainder.
– André 3000
Dec 3 at 4:02
@André3000 I think your method is just an elaboration of Rolf Hoyer's comment. It would be better if it were made into an answer, though.
– Brahadeesh
Dec 3 at 4:15
2
@RolfHoyer It would be great if you or André3000 can turn your (respective) comments into an answer.
– Brahadeesh
Dec 3 at 4:16
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Q: Find an odd prime $p$ for which $x-2$ is a divisor of $x^4 + x^3 + 3x^2 + x + 1$ in $mathbb{Z}_p[x]$.
I would rather not go through the Euclidean Algorithm for every mod $p$. Is there another way to find this particular prime?
Any help would be greatly appreciated.
abstract-algebra polynomials prime-numbers factoring
edited Dec 3 at 4:00
André 3000
12.2k22041
asked Dec 3 at 2:11
Raul Quintanilla Jr.
552
Q: Find an odd prime $p$ for which $x-2$ is a divisor of $x^4 + x^3 + 3x^2 + x + 1$ in $mathbb{Z}_p[x]$.
I would rather not go through the Euclidean Algorithm for every mod $p$. Is there another way to find this particular prime?
Any help would be greatly appreciated.
abstract-algebra polynomials prime-numbers factoring
abstract-algebra polynomials prime-numbers factoring
edited Dec 3 at 4:00
André 3000
12.2k22041
asked Dec 3 at 2:11
Raul Quintanilla Jr.
552
edited Dec 3 at 4:00
André 3000
12.2k22041
asked Dec 3 at 2:11
Raul Quintanilla Jr.
552
edited Dec 3 at 4:00
André 3000
12.2k22041
edited Dec 3 at 4:00
André 3000
12.2k22041
edited Dec 3 at 4:00
André 3000
12.2k22041
12.2k22041
asked Dec 3 at 2:11
Raul Quintanilla Jr.
552
asked Dec 3 at 2:11
Raul Quintanilla Jr.
552
asked Dec 3 at 2:11
Raul Quintanilla Jr.
552
552
8
note that $x-2$ divides $f(x)$ specifically when $f(2) = 0$, so plug in $x=2$ and find a prime divisor of the result.
– Rolf Hoyer
Dec 3 at 2:14
Rolf Hoyer has given a nice trick, but even doing this problem the "straightforward way" is quite short. We perform one polynomial division with remainder in $mathbb{Z}[x]$. The remainder will be a constant since the divisor has degree $1$, and then we pick a prime dividing the remainder.
– André 3000
Dec 3 at 4:02
@André3000 I think your method is just an elaboration of Rolf Hoyer's comment. It would be better if it were made into an answer, though.
– Brahadeesh
Dec 3 at 4:15
2
@RolfHoyer It would be great if you or André3000 can turn your (respective) comments into an answer.
– Brahadeesh
Dec 3 at 4:16
add a comment |
8
note that $x-2$ divides $f(x)$ specifically when $f(2) = 0$, so plug in $x=2$ and find a prime divisor of the result.
– Rolf Hoyer
Dec 3 at 2:14
Rolf Hoyer has given a nice trick, but even doing this problem the "straightforward way" is quite short. We perform one polynomial division with remainder in $mathbb{Z}[x]$. The remainder will be a constant since the divisor has degree $1$, and then we pick a prime dividing the remainder.
– André 3000
Dec 3 at 4:02
@André3000 I think your method is just an elaboration of Rolf Hoyer's comment. It would be better if it were made into an answer, though.
– Brahadeesh
Dec 3 at 4:15
2
@RolfHoyer It would be great if you or André3000 can turn your (respective) comments into an answer.
– Brahadeesh
Dec 3 at 4:16
8
8
note that $x-2$ divides $f(x)$ specifically when $f(2) = 0$, so plug in $x=2$ and find a prime divisor of the result.
– Rolf Hoyer
Dec 3 at 2:14
note that $x-2$ divides $f(x)$ specifically when $f(2) = 0$, so plug in $x=2$ and find a prime divisor of the result.
– Rolf Hoyer
Dec 3 at 2:14
Rolf Hoyer has given a nice trick, but even doing this problem the "straightforward way" is quite short. We perform one polynomial division with remainder in $mathbb{Z}[x]$. The remainder will be a constant since the divisor has degree $1$, and then we pick a prime dividing the remainder.
– André 3000
Dec 3 at 4:02
Rolf Hoyer has given a nice trick, but even doing this problem the "straightforward way" is quite short. We perform one polynomial division with remainder in $mathbb{Z}[x]$. The remainder will be a constant since the divisor has degree $1$, and then we pick a prime dividing the remainder.
– André 3000
Dec 3 at 4:02
@André3000 I think your method is just an elaboration of Rolf Hoyer's comment. It would be better if it were made into an answer, though.
– Brahadeesh
Dec 3 at 4:15
@André3000 I think your method is just an elaboration of Rolf Hoyer's comment. It would be better if it were made into an answer, though.
– Brahadeesh
Dec 3 at 4:15
2
2
@RolfHoyer It would be great if you or André3000 can turn your (respective) comments into an answer.
– Brahadeesh
Dec 3 at 4:16
@RolfHoyer It would be great if you or André3000 can turn your (respective) comments into an answer.
– Brahadeesh
Dec 3 at 4:16
add a comment |
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8
note that $x-2$ divides $f(x)$ specifically when $f(2) = 0$, so plug in $x=2$ and find a prime divisor of the result.
– Rolf Hoyer
Dec 3 at 2:14
Rolf Hoyer has given a nice trick, but even doing this problem the "straightforward way" is quite short. We perform one polynomial division with remainder in $mathbb{Z}[x]$. The remainder will be a constant since the divisor has degree $1$, and then we pick a prime dividing the remainder.
– André 3000
Dec 3 at 4:02
@André3000 I think your method is just an elaboration of Rolf Hoyer's comment. It would be better if it were made into an answer, though.
– Brahadeesh
Dec 3 at 4:15
2
@RolfHoyer It would be great if you or André3000 can turn your (respective) comments into an answer.
– Brahadeesh
Dec 3 at 4:16