Challenge to the community, Squaring of polynomials conjecture

[checkitagain]

Given polynomials of degree n > 2, such that they have the form of


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.


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


In \ \ terms \ \ of \ \ n, \ \ what \ \ is \ \ the \ \ greatest \ \ number \ \ of \ \ terms \ \ with


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


Is \ \ it \ \ (n + 1) \ \ terms \ ?




\text{Examples:}


(x^2 + 2x - 2)^2 \ = \ x^4 + 4x^3 - 8x + 4


(x^3 + 2x^2 - 2x + 4)^2 = x^6 + 4x^5 + 20x^2 - 16x + 16

 

[Simon Bridge]

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".

 

[checkitagain]

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".

No, if you read it again, none 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 [p(x)]^2 itself has the most terms
possible with coefficients that are zero.



\text{ * * * Note: This problem is still open.}

 

[Simon Bridge]

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.

 

[Office_Shredder]

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 x^n/2 and higher terms, then every coefficient in the square that involves x^3n/2 or a higher order term can be expressed as a linear combination of the unfixed coefficients (because to get a x^3n/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.

 

[checkitagain]

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< <?

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

(x^4 + 2x^3 - 2x^2 + 4x + 4)^2 = x^8 + 4x^7 + 28x^4 + 32x + 16


But now, the number of nonzero terms is (n + 1) = 5 (instead of a predicted 6).

 

[guillefix]

In \ \ terms \ \ of \ \ n, \ \ what \ \ is \ \ the \ \ greatest \ \ number \ \ of \ \ terms \ \ with


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

<br /> <br /> <blockquote data-attributes="" data-quote="checkitagain" data-source="post: 3675953" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> checkitagain said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> In my examples:<br /> <br /> Wait, I noticed instead that for n = 2, the number of nonzero terms is 4, or (n + 2). </div> </div> </blockquote><br /> 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&#039;t, which one are you looking for an expression?

 

[micromass]

Checkitagain, could you PM your solution to a mentor so this challenge can stay open??

 

[checkitagain]

I am asking this to be closed.

I PMed a mentor for that.

 

[micromass]

Closed by request of the OP.