Square of a polynomial

Question

Find a polynomial with more than one nonzero term such that its square has exactly same number of terms as the original polynomial.

Attempts-I tried to use variables for the polynomial and equate some to $0$. I also found that it is not possible for degree $1,2,3$. But I could not find a way how to find such a polynomial. Source - EJ Barbeau Polynomial.

consider $y=x$, $y^2 = x^2$ still the same number of terms or is this not what you meant? — The Integrator, May 15 '18 at 17:40
When the polynomial is squared and the number of non zero terms is same as the original polynomial — Brilli, May 15 '18 at 17:41
@abhishekchaudhary: That is a power series, not a polynomial. — Clayton, May 15 '18 at 17:42
There have been a little bit of change in the problem as the statement says it has more than one non zero terms — Brilli, May 15 '18 at 17:44
@abhishekchaudhary: I agree (mostly), but the problem is that you aren't starting with a polynomial. A power series squared is still a power series. — Clayton, May 15 '18 at 17:44
@clayton i just wanted to have a definition of the number of terms... The question uses some variables and wanted to equate some to 0, so the question came with the radical try to equate them all to 0.. — dan_fulea, May 15 '18 at 17:44
Please clarify your question, Brilli, as an edit to your actual question post, and please, answer @WillJagy's answer, too? — amWhy, May 15 '18 at 17:46
Example- number of terms in x^2+x^3 is 2 but in x^2+x^3+1 is 3. — Brilli, May 15 '18 at 17:46
Please clarify your question, Brilli, as *an edit to your actual question post (above), instead of offering a moving target of questions in the comments. And also share there where you encountered the problem.* Most readers who might want to answer your question are not going to look for clarification in the comments. Hence, it is your job to adjust the actual question post to match what you are telling us in the comments. — amWhy, May 15 '18 at 17:48
http://poincare.matf.bg.ac.rs/~zarkom/Polynomials_EJBarbeau.pdf — Will Jagy, May 15 '18 at 17:51

score 3 · Answer 1 · answered May 15 '18 at 17:51

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$$\left( {x}^{4}+\sqrt{2}\; {x}^{3}-{x}^{2}+\sqrt {2}\;x+1 \right) ^{2}={x} ^{8}+2\,\sqrt {2}\;{x}^{7}+7\,{x}^{4}+2\,\sqrt {2}\;x+1 $$

answered May 15 '18 at 17:51

Robert Israel

448,999

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Robert Israel could you tell how you got to the answer?? – Brilli May 15 '18 at 17:52
Set the $x^2, x^3, x^5, x^6$ coefficients to $0$ in $(x^4 +c_3 x^3 + c_2 x^2 + c_1 x+1)^2$, and solve. – Robert Israel May 15 '18 at 17:54
Thank you. I was trying too that way. I thought there would be an alternative. – Brilli May 15 '18 at 17:56
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Did you solve this with a computer or is there a natural reason to think that one should start with a $4$th degree polynomial? (I don't have any intuition that says $4$th degree should work, other than just cranking out the formulas). – Clayton May 15 '18 at 17:57
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I believed Brilli's statement that degrees $1,2,3$ don't work. $4$ is the next one to try. – Robert Israel May 15 '18 at 17:58
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If one prefers to get rid of the $\sqrt 2$'s, $(4x^4+4x^3-2x^2+2x+1)^2=16x^8+32x^7+28x^4+4x+1)$ – Hagen von Eitzen May 15 '18 at 17:58
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@RobertIsrael: I see, haha. I thought you may have had some incredible intuition. Thanks for responding. – Clayton May 15 '18 at 17:59
@RobertIsrael Is there a particular reason you chose to set those particular coefficients to $0$, as opposed to a different selection? – B. Mehta May 16 '18 at 00:04
Sometimes symmetry is your friend. – Robert Israel May 16 '18 at 02:32
But that means for any degree there is only one such polynomial. – Brilli May 18 '18 at 12:15
What means that, and why? – Robert Israel May 18 '18 at 12:54

Square of a polynomial

1 Answers1