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Proving a polynomial has no integer solution

  1. Jun 6, 2017 #1
    1. The problem statement, all variables and given/known data
    let p(x) be a polynomial with integer coefficients satisfying p(0) = p(1) = 1999
    show that p has no integer zeros

    2. Relevant equations


    3. The attempt at a solution
    ##
    p(x) = \sum_{i= 0}^{n}{a_i x^i}
    ##

    using the given information
    a0 = 1999( a prime number)
    and
    ##
    a_n + a_{n-1} ... a_1 = 0
    ##

    then rewriting p(x)

    ##
    p(x) = a_n(x_n + \frac{a_{n-1}}{a_n} x^{n-1} ... \frac{1999}{a_n})\\
    p(x) = a_n(x-r_1)(x - r_2) ....(x - r_n)\\
    ##
    i am hoping to do the rest of the proving by contradiction
    if i assume the polynomial has integer solution then how can i disprove it
     
  2. jcsd
  3. Jun 6, 2017 #2
    Use Rational roots theorem.

    If ##p/q## is a root of the given polynomial then ##p## must be a factor of ##a_0##.
    Which means ## p = 1## or ##p = 1999##.
     
  4. Jun 6, 2017 #3
    ya i tried doing that what do i get from that
     
  5. Jun 6, 2017 #4
    You know that ##q## is a factor of ##a_n##. If ##p/q## is a integer then what condition do you get on ##q## for ##p = 1## ?
     
  6. Jun 6, 2017 #5
    ##
    P(x) = (qx - p)(Q(x))\\

    q|a_n \\
    p|1999\\
    p = \pm 1 or \pm 1999\\
    ##

    if p = ±1 then q must also equal ±1
    if p = ±1999 then q must equal ±1 or ±1999
     
  7. Jun 6, 2017 #6
    So what the possible integer values ?
     
  8. Jun 6, 2017 #7
    you mean the roots the possible integer roots are ±1 and ±1999
     
  9. Jun 6, 2017 #8
    Sorry. Yes I mean roots.
     
  10. Jun 6, 2017 #9
    you mean the roots the possible integer roots are ±1 and ±1999
     
  11. Jun 6, 2017 #10
    ##1## is out because ##p(1) = 1999##. Can you eliminate others ?
     
  12. Jun 6, 2017 #11
    i am sorry i just cant think of ways to eliminate the rest give me another hint
    edit:
    after much trying managed to eliminate just 1 of the roots
    if we a
    ##
    a_n + a_{n- 1} ... a_1 = 0\\
    a_n + a_{n-1} .... a_2 = -a_1\\
    p(-1) = a_n(-1)^n + a_{n-1} (-1)^n-1 .... -a_1 + 1999\\
    p(-1) = a_n(-1)^n + a_{n-1} (-1)^n-1 .... a_2 +a_2 +a_3 + .... a_n + 1999\\
    ##
    the right hand side will cancel out all the terms with odd subscripts leaving only 2 of each even subscript terms and since they are integers the right hand side is 1 mod 2 thus cannot be zero is this correct?
    please give me hints for the next two roots
     
    Last edited: Jun 6, 2017
  13. Jun 6, 2017 #12

    Yes except I think it should be ##p(-1) = a_n(-1)^n + a_{n-1} (-1)^\color{green}{n-1} .... a_2 +a_2 +a_3 + .... a_n + 1999##.

    Hint :

    Product of roots is ##|a_0/a_n|##

    Or ##\dfrac{|a_0|}{|a_n|} = |r_1r_2 \cdots r_n|##

    If one root is ##1999##
    then,
    ##|1999| = |a_n||r_2 \cdots r_n| \times 1999 \implies 1 = |a_n||r_2 \cdots r_n|##
     
  14. Jun 6, 2017 #13

    haruspex

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    If 1999 is a root, how many times might 1999 divide ai1999i? Given that it divides a0 exactly once, what does that tell you about a1? Then, what about the relationship between a2 and a3?
     
    Last edited: Jun 6, 2017
  15. Jun 6, 2017 #14
    @haruspex
    i think what @Buffu wrote is correct as he substituted ## - a_1 = a_2 + a_3 ... a_n##

    @Buffu
    following on what you said
    ## 1 = (\mid a_n \mid)(\mid r_2 r_3 ...r_n) ## since ##a_n ## is integer## \mid a_n \mid = 1 ##
    then this make ## p(x) ## a monic polynomial then its roots can only be integers or irrationals not fractions then from ##\mid r_2 r_3 ...r_n\mid = 1## we get that ##r_i## has to equal 1 but its shown that 1 is not possible hence 1999 is not possible is this valid ?
     
  16. Jun 6, 2017 #15

    haruspex

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    Ok, I misinterpreted the ......

    I'm not following. If ##\mid r_2 r_3 ...r_n\mid=\frac 12 ## then ##(\mid a_n \mid)=2##, etc. How are you ruling that out?
     
  17. Jun 7, 2017 #16
    oh ya i got over excited and did that and i am stuck again one more hint please is there some kind of restriction on an
    should i be writing ## \mid a_n \mid ## as ## \mid a_1 + a_2 ... a_{n-1} \mid ## does that help
     
    Last edited: Jun 7, 2017
  18. Jun 7, 2017 #17

    haruspex

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    Did you think about my post #13?
     
  19. Jun 7, 2017 #18
    oh i completely didn't see the post only noticed it after you pointed it out don't know what happened there will take a look now thanks oh wait you edited i forgot to see the edit
     
  20. Jun 7, 2017 #19
    @ haruspex
    are you hinting at geometric series
    because i thought of it at first but the fact that a2 + ....an = 0
    then i considered all the terms in that to be same but alternating signs but that forces n to be odd this was initial line of thought
    i am sorry if am not seeing what you are suggesting
     
  21. Jun 7, 2017 #20

    haruspex

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    If you plug x=1999 into the polynomial you will get a factor of 1999 in every term of the sum. Some terms will have extra factors of 1999. But in the end, the sum is zero if 1999 is a root. Can there be one term with fewer factors of 1999 than the rest? What does that tell you about a1?
     
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