Let $n\in \mathbb{N}$. Prove that there exists a rational number $a_n$ for which $$(x^2+\frac{1}{2}x+1)\mid(x^{2n}+a_nx^n+1)$$
My attempt :
I try $n=2$,
$(x^2+\frac{1}{2}x+1)(x^2-\frac{1}{2}x+1)=x^4+\frac{7}{4}x^2+1$
$a_2 = \frac{7}{4}$
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1Can you imagine there are two votes to close the question as "off topic"?! I quote what is certainly not off-topic, from the Help Center: "Understanding mathematical concepts and theorems, Mathematical problems such as one might come across in a course or textbook, Solving mathematical puzzles." – Jul 29 '17 at 16:24
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I don't understand why this is off topic. – user403160 Jul 29 '17 at 16:39
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1It isn't, of course. Just some people at this site are very, very sick, and often forget to take their meds. They can vote to close only with a reputation of 3000 or more, so they know enough math to realize that this is not off-topic. They're just misusing the power of their reputation, the poor bastards. Don't take it personally. – Jul 29 '17 at 23:20
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@Professor Vector. Thank you :) – user403160 Jul 30 '17 at 03:02
2 Answers
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just an idea
The roots of the left polynom are
$$x_1=\frac{-1+i\sqrt {15} }{4}=e^{it}$$ $$x_2=\frac {-1-i\sqrt {15} }{4}=e^{-it} $$
they are also roots of the right one
$$e^{2int}+a_ne^{int}+1=0$$ $$e^{-2int}+a_ne^{-int}+1=0$$
thus $$a_n=-\frac {\sin (2nt)}{\sin (nt)} $$
$$=-2\cos (nt) $$
For example, $$a_2=-2\cos (2t)=2 (1-2\cos^2 (t)) $$ $$=2-4\frac {1}{16}=\frac {7}{4} $$
hamam_Abdallah
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Correct, and since $\cos t=(e^{it}+e^{-it})/2=-1/4,$ you have the explicit expression $a_n=-2 T_n(-1/4),$ where $T_n$ are the Chebyshev polynomials of first kind (https://en.wikipedia.org/wiki/Chebyshev_polynomials). – Jul 29 '17 at 15:54
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@ProfessorVector Thanks a lot Professor. I did not think about Chebyshev. – hamam_Abdallah Jul 29 '17 at 15:58
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Thank you, Salahamam_Fatima and ProfessorVector. Can this problem be solved using knowledge at high school level ? – user403160 Jul 29 '17 at 16:42
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We define $$p_n(x)=x^{2n}+a_n\,x^n+1,$$ so we have to prove $$\left.x^2+\frac12\,x+1\right|p_n(x).$$ This is clearly true for $n=1$ with $a_1=1/2$, and it was shown for $n=2$ with $a_2=7/4$. Induction step: we assume it's true for $n=k-1$ and $n=k$. Then, as can be seen by expanding the products, $$p_{k+1}(x)=(x^{2k}+1)\left(x^2+\frac12\,x+1\right)-\frac12\,x\,p_k(x)-x^2\,p_{k-1}(x),$$ if only $$a_{k+1}=-\frac12\,a_k-a_{k-1},$$ so $$\left.x^2+\frac12\,x+1\right|p_{k+1}(x).$$
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Since $a_1, a_2 \in \mathbb{Q}$, so $a_n \in \mathbb{Q}, ;\forall n \in \mathbb{N}$. Thank you very much, Professor Vector. – user403160 Jul 30 '17 at 03:14