1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Proving a polynomial cannot be factored with integer coefficients

  1. Jun 16, 2017 #1
    1. The problem statement, all variables and given/known data
    upload_2017-6-16_20-44-17.png



    2. Relevant equations


    3. The attempt at a solution
    i tried to do it by writing it as
    ##
    a_{1999} x^{1999} + a_{1998} x^{1998} ... a_0 \pm1 = 0
    ##
    for 1999 different integer values of x
    i am thinking of writing it as
    ##
    a_{1999} x^{1999} = -a_{1998} x^{1998} - a_{1997} x ^ {1997} .... a_0 \pm1
    ##
    does this lend any information

    i also tried to graph it
    a 1999 degree polynomial has 1998 stationary points and at most 2000 points at which it is plus minus 1
     
    Last edited: Jun 16, 2017
  2. jcsd
  3. Jun 16, 2017 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    I would assume that you can factor it, and then look for a contradiction. What can you say about the value of the factors at these points?

    Technical detail: The problem statement should say "non-constant polynomials", otherwise it is wrong: you can write p(x) as 1*p(x).
     
  4. Jun 16, 2017 #3

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Use braces: in TeX/LaTeX, if you have a subscript or superscript consisting of more than one character, you need to put it in braces, like this: a_{1999} and x^{1999}. Without braces, a_1999 and x^1999 give you get this: ##a_1999, x^1999##; but with braces you get this: ##a_{1999}, x^{1999}##.
     
  5. Jun 16, 2017 #4
    @mfb
    sure give me some time i will reply soon
    @Ray Vickson
    sorry about that still new to letex thanks edited it forgot to check after posting
     
  6. Jun 17, 2017 #5
    @mfb
    this is what i did
    assuming it can be factored into a product of two polynomials with integer coeffecients it can be written as
    ##
    (a_n x^ n + a_{n-1} x^{n-1} .... a_0)(b_{1999-n}x^{1999-n} + b_{1998-n} x^{1998-n} .... b_0)\\
    where \\
    a_i , b_i = integers
    ##
    hence
    ##
    (a_n x^ n + a_{n-1} x^{n-1} .... a_0) = \pm 1\\

    (b_{1999-n}x^{1999-n} + b_{1998-n} x^{1998-n} .... b_0) = \pm 1
    ##
    for 1999 different values of integer x
    if n is greater than or equal to 2
    then the degree of the b polynomial is less than 1998 and cannot be plus minus 1 for 1999 distinct values
    but if n = 1 the degree of a polynomial is just 1 and certainly can at most be plus minus 1 2 times
    hence it cannot be factored like this
    is this correct?
     
  7. Jun 17, 2017 #6

    fresh_42

    Staff: Mentor

    But you have two polynomials: ##a(x) + 1## and ##a(x) - 1##. Can't the ##1999## many zeros be spread on these two?
     
  8. Jun 17, 2017 #7
    @fresh_42
    i dont quite get why you are talking about zeros here
    the question states the initial polynomial is ±1 for 1999 different integer values of x and since both of the a and b polynomial have only integer terms in them their absolute value must equal 1 also for 1999 values of x

    oh wait now that i actually drawn it out i realise i am wrong
    because a cubic graph can have 4 points with absolute value equal to 1 and a pentic graph has 8 such points and hence i conjectured for n th degree there will be 2n - 2 such points
    so 1998 degree has 3994 such points not 1999 as i initially thought
    and now that i think more
    1000 degree can have 1998 such points so the b polynomial must be at least of at least 1001 degree and hence a becomes of degree 998 degree which is not possible now correct? i am really hesitant about this method as it seems very vague to me myself but that is the only one i can think of please help
     
  9. Jun 17, 2017 #8
    no sorry once again after further checking my conjecture was wrong
    it turns out that degree 3 graph has 6 such points and degree 5 graph has 10 such points so i now conjecture that number of such points is 2n
    so the b polynomial must be atleast 1000 degree now then the a polynomial becomes 999 degree which only has 1998 such points which is not possible
     
  10. Jun 17, 2017 #9

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Correct.
     
  11. Jun 17, 2017 #10
    really wow thanks
     
  12. Jun 18, 2017 #11
    can i ask a similar question about factoring polynomials here
     
  13. Jun 18, 2017 #12

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    If it is a different problem, please open a new thread.
     
  14. Jun 18, 2017 #13
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Proving a polynomial cannot be factored with integer coefficients
  1. Factoring Polynomials (Replies: 2)

  2. Factoring Polynomial (Replies: 4)

  3. Factoring a polynomial (Replies: 4)

  4. Polynomial factoring (Replies: 7)

Loading...