There exists rational number $a_n$, $(x^2+\frac{1}{2}x+1)\mid(x^{2n}+a_nx^n+1)$ 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 $n=2$, 
$(x^2+\frac{1}{2}x+1)(x^2-\frac{1}{2}x+1)=x^4+\frac{7}{4}x^2+1$
$a_2 = \frac{7}{4}$

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,
$$a_2=-2\cos (2t)=2 (1-2\cos^2 (t)) $$
$$=2-4\frac {1}{16}=\frac {7}{4} $$
