Results 1 to 15 of 15

Math Help - Modular arithmetic - Polynomials

  1. #1
    Member
    Joined
    Nov 2010
    Posts
    164

    Modular arithmetic - Polynomials

    Sorry for bothering again, I have thee questions. I did the first two ones and just want a verification of my answers and guidance for solving the third question.

    1) Find all the solutions of x^2+2x+4=0 in Z/6Z

    Solution: Let f(x)=x^2+2x+4=0 then only f(2)=12 is congruent to 0 (mod 6) so x=2 is the only solution.

    2) Find all prime numbers p such that x+2 is a factor of x^4+x^3+x^2-x+1 in Fp[x]

    Solution: Let f(x)=x^4+x^3+x^2-x+1. If x+2 is a factor of f(x) then f(2) is congruent to 0 (mod p), p a prime number

    f(2)=27 so we must find the prime numbers that divide 27, so p=3 is the only solution

    3) Find all the positive integers n such that x^2+3 divides x^5-10x+12 in (Z/nZ) [x]


    If someone can check my answers for question 1 and 2 and give me an answer for question 3 I would appreciate it. Thanks in advance!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member TheChaz's Avatar
    Joined
    Nov 2010
    From
    Northwest Arkansas
    Posts
    600
    Thanks
    2
    1. f(2) = 12 = 0, so 2 is a solution, but that might not be enough (for your teacher, at least) to just say "this is the only solution".
    You could just show your work for f(0), f(1), f(3), f(4), and f(5). (Hopefully that will add about 4 lines!)

    2. Uh oh! When (x + 2) is a factor, then f(-2) = 0 !!! Rework this one, with that slight change.

    3. Have you tried long division?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2010
    Posts
    164
    Thank you. For qn 1 I showed my working, for question 2 f(-2)=15 so p=3, p=5.
    For qn 3 how to do a long division reducing modulo n if we don't know n? Thanks again!

    Ok, I did a long division and found that x^5-10x+12=(x^2+3)*(x^3-3x)-x+12, i.e my remainder is -x+12
    How can I proceed from that? Thank you again!
    Last edited by Darkprince; March 14th 2011 at 09:46 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member TheChaz's Avatar
    Joined
    Nov 2010
    From
    Northwest Arkansas
    Posts
    600
    Thanks
    2
    First you do the long division (!).
    Then the remainder (which Wolfram claims to be 12 - x) should equal/be congruent to zero. Then... you're almost there!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Nov 2010
    Posts
    164
    So x=12 implies n=12 is the only poaitive integer n such that x^2+3 divides x^5-10x+12??
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member TheChaz's Avatar
    Joined
    Nov 2010
    From
    Northwest Arkansas
    Posts
    600
    Thanks
    2
    We are solving 12 = 0 (mod n) for n > 0
    So how about the factors of 12???
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,845
    Thanks
    320
    Awards
    1
    Quote Originally Posted by Darkprince View Post
    Sorry for bothering again, I have thee questions. I did the first two ones and just want a verification of my answers and guidance for solving the third question.

    1) Find all the solutions of x^2+2x+4=0 in Z/6Z

    Solution: Let f(x)=x^2+2x+4=0 then only f(2)=12 is congruent to 0 (mod 6) so x=2 is the only solution.

    2) Find all prime numbers p such that x+2 is a factor of x^4+x^3+x^2-x+1 in Fp[x]

    Solution: Let f(x)=x^4+x^3+x^2-x+1. If x+2 is a factor of f(x) then f(2) is congruent to 0 (mod p), p a prime number

    f(2)=27 so we must find the prime numbers that divide 27, so p=3 is the only solution

    3) Find all the positive integers n such that x^2+3 divides x^5-10x+12 in (Z/nZ) [x]


    If someone can check my answers for question 1 and 2 and give me an answer for question 3 I would appreciate it. Thanks in advance!
    The first two looks good to me.

    For the third I would make the division:
    \displaystyle \frac{x^5 - 10x + 12}{x^2 - 3} = x^3 - 3x - \frac{x - 12}{x^2 + 3}

    For what x (mod n) is x - 12 = 0?

    -Dan

    Wow. A lot happened as I was writing this!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,682
    Thanks
    614
    Hello, Darkprince!

    I have another approach to #1 . . .


    \text{(1) Find all the solutions of: }\, x^2+2x+4\:\equiv\:0\:\text{ (mod 6)}

    \text{Since }2\:\equiv\:-4\text{ (mod 6)}

    . . \text{the equation becomes: }\:x^2 - 4x + 4 \:\equiv\:0\text{ (mod 6)}


    \text{Then we have: }\:(x-2)^2 \:\equiv\:0\text{ (mod 6)} \quad\Rightarrow\quad x-2 \:\equiv\:0\text{ (mod 6)}


    . . \text{Therefore: }\:x \:\equiv\:2\text{ (mod 6)}

    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Nov 2010
    Posts
    164
    Thanks but why are we solving 12 = 0 (mod n) for n > 0? I just don't see the point, sorry for asking again!
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,845
    Thanks
    320
    Awards
    1
    Quote Originally Posted by Darkprince View Post
    Thanks but why are we solving 12 = 0 (mod n) for n > 0? I just don't see the point, sorry for asking again!
    The remainder of the long division has the numerator x - 12 for any n. The only way to make the remainder 0 then, is to let x - 12 = 0 (mod n).

    -Dan
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member TheChaz's Avatar
    Joined
    Nov 2010
    From
    Northwest Arkansas
    Posts
    600
    Thanks
    2
    Quote Originally Posted by topsquark View Post
    ...
    Wow. A lot happened as I was writing this!
    We also solved the general quintic. Where were you?!?

    But seriously, forks...

    We want the remainder to "equal" zero.
    The remainder is (congruent to) 12. So we want 12 to "equal" zero.
    12 = 0 (mod 1) ... in a lame ("trivial") kind of way
    12 = 0 (mod 2)
    12 = 0 (mod 3)
    12 = 0 (mod ....)

    Are we making headway?

    (edit: apparently a lot happened while *I* was writing! )
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,845
    Thanks
    320
    Awards
    1
    Quote Originally Posted by TheChaz View Post
    We also solved the general quintic. Where were you?!?
    Coming up with a more intuitive method for solving the four color theorem.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Member
    Joined
    Nov 2010
    Posts
    164
    Thank you all for the answers. I am yet trying to understand the logic behind this. The remainder is -x+12 for any n. But we want a zero remainder, so -x+12=0 =>x=12. After that I am a bit confused! Thanks again for any help!!
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,845
    Thanks
    320
    Awards
    1
    Quote Originally Posted by Darkprince View Post
    Thank you all for the answers. I am yet trying to understand the logic behind this. The remainder is -x+12 for any n. But we want a zero remainder, so -x+12=0 =>x=12. After that I am a bit confused! Thanks again for any help!!
    Again, we are working in mod n, so we need x = 12 (mod n). Is 12 the only solution for x? (For an example, what if n = 2?)

    -Dan
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Member
    Joined
    Nov 2010
    Posts
    164
    n can be 1,2,3,4,6,12 right? Thanks for the help guys I think I figured it out
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. modular arithmetic help
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: May 9th 2011, 05:50 AM
  2. Modular arithmetic
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: May 3rd 2011, 11:07 PM
  3. Modular arithmetic.
    Posted in the Number Theory Forum
    Replies: 4
    Last Post: March 10th 2011, 05:21 PM
  4. Modular arithmetic
    Posted in the Number Theory Forum
    Replies: 6
    Last Post: June 29th 2010, 07:51 PM
  5. Modular Arithmetic
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: March 28th 2010, 05:08 AM

Search Tags


/mathhelpforum @mathhelpforum