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Given that $x^2 + x - 1 = 0$, what is $$V \equiv \frac{x^{10} + x^8 + x^2 + 1}{x^{10} + x^6 + x^4 + 1} = \; ?$$

I have reduced $V$ to $\dfrac{x^8 + 1}{(x^4 + 1) (x^4 - x^2 + 1)}$, if you would like to know.

Bumblebee
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hengxin
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5 Answers5

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HINT: Note that $$\color{Green}{x^2=-x+1}.$$ By multiplying the given expression by $x,$ we can obtained $$\color{Green}{x^3=2x-1}.$$ Again by multiplying $x$ we have $$\color{Green}{x^4=-3x+2}$$ and so on..

Bumblebee
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$$x^2=1-x$$ $$x^4=(1-x)^2=x^2-2x+1=1-x-2x+1=2-3x$$ $$x^6=(1-x)(2-3x)=3x^2-5x+2=3(1-x)-5x+2=5-8x$$ $$x^8=(2-3x)^2=9x^2-12x+4=9(1-x)-12x+4=13-21x$$ $$x^{10}=(1-x)(13-21x)=21x^2-34x+13=21(1-x)-34x+13=34-55x$$ Therefore, $$x^{10}+x^8+x^2+1=34-55x+13-21x+1-x+1=49-77x$$ $$x^{10}+x^6+x^4+1=34-55x+5-8x+2-3x+1=42-66x$$ Your equation =$$\frac{49-77x}{42-66x}=\frac{7}{6}$$

Kay K.
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    It might be worth noting here that the appearance of Fibonacci numbers in converting powers of $x$ to linear polynomials in $x$ is Not A Coincidence... – Steven Stadnicki Nov 23 '15 at 06:36
  • @StevenStadnicki An explanation? Or another answer? – hengxin Nov 23 '15 at 06:43
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    @hengxin If $y=-x$ then it's easy to see that $y$ satisfies $y^2=y+1$; and if we define $a_n$ and $b_n$ by $y^n=a_ny+b_n$ then multiplying gives $y^{n+1} = a_ny^2+b_ny$ = $(a_n+b_n)y+a_n$; in other words, $b_{n+1} = a_n$ (or, shifting indices, $b_n=a_{n-1}$) and $a_{n+1}=a_n+b_n=a_n+a_{n-1}$ - so $a_n$ satisfies the Fibonacci recurrence. – Steven Stadnicki Nov 23 '15 at 07:03
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Applying polynomial long division, we have

$\begin{array}{rlllllllllll} &~~1x^8-1x^7+3x^6+\dots\\ \hline x^2+x-1&|x^{10}+0x^9+x^8+0x^7+0x^6+0x^5+0x^4+0x^3+1x^2+0x+1\\ &~x^{10}+x^9-x^8\\ \hline &~~~~~-x^9+2x^8+0x^7\\ &~~~~~-x^9-x^8+x^7\\ \hline &~~~~~~~~~~~~~~~~~3x^8+\dots\\ &~~~~~~~~~~~~~~~~~~\vdots \end{array}$

eventually arriving at $$x^{10}+x^8+x^2+1\\=(x^8-x^7+3x^6-4x^5+7x^4-11x^3+18x^2-29x+48)(x^2+x-1)+49-77x\\=49-77x$$

Similarly, applying polynomial long division to the denominator, it simplifies to

$42-66x$.

So the fraction in general simplifies to

$\frac{49-77x}{42-66x}=\frac{7}{6}$

JMoravitz
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$$\frac{x^{10} + x^8 + x^2 + 1}{x^{10} + x^6 + x^4 + 1} =\dfrac{(x^2+1)(x^8+1)}{(x^4+1)(x^6+1)}$$

$$=\dfrac{\left(x+\dfrac1x\right)\left(x^4+\dfrac1{x^4}\right)}{\left(x^2+\dfrac1{x^2}\right)\left(x^3+\dfrac1{x^3}\right)}$$

$$=\dfrac{\left(x^2+\dfrac1{x^2}\right)^2-2}{\left(x^2+\dfrac1{x^2}\right)\left(x^2+\dfrac1{x^2}+1\right)}$$

Now $x^2+x-1=0\implies x-\dfrac1x=-1$ as $x\ne0$

Now $x^2+\dfrac1{x^2}=\left(x-\dfrac1x\right)^2+2=1+2$

Can you take it from here?

lab bhattacharjee
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Another approach is to note the following

$$\begin{align} x^2&=1-x\\ x^4&=(1-x)^2=1-2x+1-x&=2-3x\\ x^6&=(2-3x)(1-x)=2-5x+3(1-x)&=5-8x\\ x^8&=(5-8x)(1-x)=5-13x+8(1-x)&=13-21x\\ x^{10}&=(13-21x)(1-x)=13-34x+21(1-x)&=34-55x \end{align}$$

Our fraction becomes

$${49-77x\over 42-66x}={7\over 6}$$

marwalix
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  • You seem to have an arithmetic mistake starting on line three. $(2-3x)(1-x)=2-5x\color{Red}{+}3x^2 = 2-5x+3-3x=5-8x$ – JMoravitz Nov 23 '15 at 06:29