Matematické Fórum

stuart clark · 29. 04. 2011 08:31

does there exist a positive integer $n\geq 2$ so that $f(x) =1+4x+4x^2+.....+4x^{2n}$ is a perfect square polynomial

musixx · 29. 04. 2011 11:16 — Editoval musixx (29. 04. 2011 11:41)

Since f(x) must be a perfect square polynomial, then all its roots must be multiple roots. In the same time, all those roots must be roots of f'(x) with a well-known consequency to their multiplicities.

Degree of f'(x) is 2n-1, and if f'(x) contained two distinct roots, then a degree of f(x) would be bigger than 2n -- a contradiction.

$f^\prime(x)=8nx^{2n-1}+\cdots+4$ .

Therefore f'(x) must be of a form $8n(x-x_0)^{2n-1}$ with the unique real root (consider complex conjugates to realize that the root must be a real one), and comparing the absolute terms, we have the only candidate $x_0=-\sqrt[2n-1]{\frac1{2n}}$ .

It is easy to see (verify by yourself) that since the absolute term of $f(x)=4$x+\sqrt[2n-1]{\frac1{2n}}$^{2n}$ is $4$\frac1{2n}$^{\frac{2n}{2n-1}}$ , then it is equal to 1 only if n=1.

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Can I ask who are you? You publish here more advanced questions from analysis and algebra, moreover in English which is not so common here... For sure, it is no problem. It is just my (maybe not only my) personal interest. I can remember that e.g. Marian, Rumburak, Pavel, and Pavel Brožek have answered your questions, as well.

byk7 · 29. 04. 2011 14:56

↑ musixx: He si probably from India. Am I right?

stuart clark · 03. 05. 2011 09:17

Thanks musixx for solving my question also thanks to all members to show interest on my questions.
and byk7 you are saying right I am from India.

byk7 · 03. 05. 2011 19:55

↑ stuart clark: Can I ask you, too? How/when did you find this website?

Pavel · 04. 05. 2011 22:15

Let $n\geq 2$ . Then

$f(x)&=1+4x+4x^2+\ldots+4x^{2n}=1+4x(1+x+\ldots+x^{2n-1})=1+4x\frac{1-x^{2n}}{1-x}= \\ &=1+\frac{4(x-x^{2n+1})}{1-x}\,.$

Prove that $f$-\frac 12$<0$ for any $n\in\mathbb N,\ n\geq 2$ .

$f$-\frac 12$=1+\frac{4$-\frac 12-\(-\frac 12$^{2n+1}\)}{1-$-\frac 12$}=1+\frac 23\cdot$-2+\frac 1{2^{2n-1}}$=-\frac 13+\frac 1{3\cdot 2^{2n-2}}\,.$

The sequence $\{-\frac 13+\frac 1{3\cdot 2^{2n-2}}\}_{n=2}^{\infty}$ is decreasing therefore

$f$-\frac 12$=-\frac 13+\frac 1{3\cdot 2^{2n-2}}\leq-\frac 13+\frac 1{3\cdot 2^{2\cdot 2-2}}=-\frac{1}{4}<0.$

However, values of a perfect square polynomial can be only nonnegative real numbers. Hence $f$ cannot be a perfect square polynomial.