Results 1 to 10 of 10

Math Help - square polynomials (mod 8 )

  1. #1
    Member
    Joined
    Sep 2006
    Posts
    83

    square polynomials (mod 8 )

    How should I prove this question?

    Prove :
    If a is any integer and the polynomial f(x) = x^2 +ax + 1 factors (polymod 8) , then f(x) is in fact a square; i.e. f(x)≡(x+c)^2 (polymod 8) for some non-negative integer c less than 8.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by beta12 View Post
    How should I prove this question?

    Prove :
    If a is any integer and the polynomial f(x) = x^2 +ax + 1 factors (polymod 8) , then f(x) is in fact a square; i.e. f(x)≡(x+c)^2 (polymod 8) for some non-negative integer c less than 8.
    It seems to me it is a direct connection to quadradic residues.
    Remember from field theory that a polynomial of degree 2 or 3 is reducible over a polynomial ring (in this case the integers under multiplication modulo 8) if and only if it has a zero.

    Thus, what we are saying is that
    x^2+ax+1≡0(mod 8)
    For some x.
    We will express this in a more efficient way,
    x^2+ax+1≡0(mod 2^3)
    That means the discriminant of this quadradic is a quadradic residue of 2.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2006
    Posts
    83
    Hi perfecthacker,

    This question wants us to find the possible values of a. i.e. for which non-negative a less than 8 does f(x) factor?

    Do you have any idea about that?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member malaygoel's Avatar
    Joined
    May 2006
    From
    India
    Posts
    648
    Quote Originally Posted by beta12 View Post
    How should I prove this question?

    Prove :
    If a is any integer and the polynomial f(x) = x^2 +ax + 1 factors (polymod 8) , then f(x) is in fact a square; i.e. f(x)≡(x+c)^2 (polymod 8) for some non-negative integer c less than 8.
    what is meant by statement:
    f(x)≡(x+c)^2 (polymod 8)

    I am missing something!

    Keep SMiling
    Malay
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by malaygoel View Post
    what is meant by statement:
    f(x)≡(x+c)^2 (polymod 8)

    I am missing something!
    That means the quadradic polynomial is equal to the square of (x+c) (under modulo 8).
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Sep 2006
    Posts
    83
    Let me write the question again clearly.

    Prove :
    If a is any integer and the polynomial f(x) = x^2 +ax + 1 factors (polymod 8) , then f(x) is in fact a square; i.e. f(x)≡(x+c)^2 (polymod 8) for some non-negative integer c less than 8.

    What are the possible values of a ? That is , for which non-negative a less than 8 does f(x) factor?

    still can't figure out the values of a!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by beta12 View Post
    Let me write the question again clearly.

    Prove :
    If a is any integer and the polynomial f(x) = x^2 +ax + 1 factors (polymod 8) , then f(x) is in fact a square; i.e. f(x)≡(x+c)^2 (polymod 8) for some non-negative integer c less than 8.

    What are the possible values of a ? That is , for which non-negative a less than 8 does f(x) factor?

    still can't figure out the values of a!
    It was clearly the first time I was lazy to answer.
    ---
    Since,
    x^2+ax+1≡(x+c)^2=x^2+2cx+c^2 (mod 8)
    By definition of polynomial equality we need that,
    a=2c
    1=c^2
    .
    So for what values of Z_8 do you have that, a number is is its own inverse. Basic computation for 0,1,...,7 shows that: 3,5,7.
    Then,
    a=2(3)=6
    a=2(5)=2
    a=2(7)=6

    So the two polynomials you have are,
    x^2+2x+1
    x^2+6x+1
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Sep 2006
    Posts
    83
    Hi perfecthacker,

    I got it now.

    Do you know how to prove this question too?

    Prove: If a is any integer and the polynomial f(x) = x^2 + ax +1 factors (polymod 9), then there are three distinct non-negative integers y less than 9 such that f(y)≡ 0 ( mod 9).
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by beta12 View Post
    Prove: If a is any integer and the polynomial f(x) = x^2 + ax +1 factors (polymod 9), then there are three distinct non-negative integers y less than 9 such that f(y)≡ 0 ( mod 9).
    It factos,
    x^2+ax+1 = (x+b)(x+c)
    Thus,
    x^2+ax+1=x^2+(b+c)x+bc
    Thus,
    a=b+c
    1=bc
    Thus, you need to find all values of 'a'
    such that you can find integers 1<=b,c<=9
    That make the above statement true.

    There should be 3 values of 'b' and 'c'.
    The first condition: a=b+c is not important.
    All you need to find is distinct 'b' and 'c' such that they are inverses (bc=1) and then declare a=b+c.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Sep 2006
    Posts
    83
    Hi perfecthacker,

    Thank you very much.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: July 2nd 2011, 08:03 PM
  2. Replies: 7
    Last Post: April 7th 2011, 01:38 PM
  3. Replies: 7
    Last Post: January 8th 2010, 04:13 AM
  4. Completing the Square (Quadratic Polynomials)
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: June 3rd 2009, 05:58 AM
  5. Replies: 5
    Last Post: November 29th 2005, 04:22 PM

Search Tags


/mathhelpforum @mathhelpforum