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Challenge to the community, Squaring of polynomials conjecture

  1. Dec 18, 2011 #1
    Given polynomials of degree n > 2, such that they have the form of


    [itex]p(x) \ = \ x^n \ + \ a_1x^{n - 1} \ + \ a_2x^{n - 2} \ + \ a_3x^{n - 3} \ + \ ... \ + \ a_{n - 2}x^2 \ + \ a_{n - 1}x \ + \ a_n.[/itex]


    [itex]And \ \ all \ \ of \ \ the \ \ a_i \ \ are \ nonzero \ integers \ (which \ you \ get \ to \ choose \ for \ each \ n).[/itex]


    [itex]In \ \ terms \ \ of \ \ n, \ \ what \ \ is \ \ the \ \ greatest \ \ number \ \ of \ \ terms \ \ with [/itex]


    [itex]\ \ coefficients \ \ that \ \ are \ \ zero \ \ that \ \ [p(x)]^2 \ \ can \ \ have \ ?[/itex]


    [itex]Is \ \ it \ \ (n + 1) \ \ terms \ ?[/itex]




    [itex]\text{Examples:}[/itex]


    [itex](x^2 + 2x - 2)^2 \ = \ x^4 + 4x^3 - 8x + 4[/itex]


    [itex](x^3 + 2x^2 - 2x + 4)^2 = x^6 + 4x^5 + 20x^2 - 16x + 16[/itex]
     
    Last edited: Dec 19, 2011
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  3. Dec 19, 2011 #2

    Simon Bridge

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    By that description, the greatest number of terms that (p(x))^2 can have with zero coefficients is 2n-1

    p(x)=x^n (a polynomial of order n with n-1 zero coefficients)

    (p(x))^2 = x^2n is a polynomial order 2n with 2n-1 coefficients.

    Of course, I have not counted a0 in that.

    Unless you mean something different by "zero coefficient".
     
  4. Dec 19, 2011 #3

    No, if you read it again, [itex]none[/itex] of the terms of p(x) are to have
    coefficients that are zero. I want to know a formula, if it exists,
    in terms of n, such that [itex][p(x)]^2 [/itex] itself has the most terms
    possible with coefficients that are zero.



    [itex]\text{ * * * Note: This problem is still open.}[/itex]
     
  5. Dec 19, 2011 #4

    Simon Bridge

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    Ah - none of the coefficients of p(x) can be zero.
    So we are trying to eliminate coefficients by making some of the initial coefficients negative.

    Enumerating coefficients - s we want to maximize N0.

    So I figure the real challenge is to find a simpler proof by limiting to the case n=2.
    Otherwise there is some crunching through matrices to do.
     
  6. Dec 19, 2011 #5

    Office_Shredder

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    Is there a particular reason you inquire if it's n+1 terms given that both your examples only end up with n-1 non-zero terms?

    As a first glance, let's just start picking the high coefficients at random and then see if we can select the low coefficients intelligently. If we fix every coefficient of xn/2 and higher terms, then every coefficient in the square that involves x3n/2 or a higher order term can be expressed as a linear combination of the unfixed coefficients (because to get a x3n/2 you must involve at least one term of degree n/2 or greater). And we have n/2 unknowns, so on average you should be able to get a solution.
     
  7. Dec 19, 2011 #6
    In my examples:

    Wait, I noticed instead that for n = 2, the number of nonzero terms is 4, or (n + 2).

    And, for n = 3, the number of nonzero terms is 5, or again (n + 2).


    Another example:


    n = 4

    [itex](x^4 + 2x^3 - 2x^2 + 4x + 4)^2 = x^8 + 4x^7 + 28x^4 + 32x + 16[/itex]


    But now, the number of nonzero terms is (n + 1) = 5 (instead of a predicted 6).
     
  8. Dec 19, 2011 #7
    Wait I got confused, so first you are asking about the number of coefficients that are zero and then about the number of coeffeficients that aren't, which one are you looking for an expression?
     
  9. Dec 19, 2011 #8

    micromass

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    Checkitagain, could you PM your solution to a mentor so this challenge can stay open??
     
  10. Dec 19, 2011 #9
    I am asking this to be closed.

    I PMed a mentor for that.
     
  11. Dec 19, 2011 #10

    micromass

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    Closed by request of the OP.
     
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