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Recently I was asked a question by my student that completely stumped me.

$$\text{If }(w + 1)(w - 1) = w\text{, find } { w }^{ 10 }+\frac { 1 }{ { w }^{ 10 } }. $$

One "cheat" method that I used was to solve for the exact value of $w$ from the given first equation, and then substitute it into the requested expression that we were asked to find. I got $123$ as the answer. However, I'm quite sure there's an algebraic way to solve this. Anyone wants to give this a ahot?

InsideOut
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Eric
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3 Answers3

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We have $w-\frac{1}{w}=1$, and then $w^2+\frac{1}{w^2}=3$. Put $u_n=w^{2n}+\frac{1}{w^{2n}}$; we have $u_0=2$, $u_1=3$, and $$u_{n+1}(w^2+\frac{1}{w^2})=u_{n+2}+u_n$$ Hence $u_{n+2}=3u_{n+1}-u_n$, it is easy to compute $u_2,u_3,u_4$, and finally $u_5$.

Kelenner
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From $w^2=w-1=0$ by dividing with $w$, we get $$w-\frac{1}{w}=1,$$ from where we get $$w^2+\frac{1}{w^2}=(w-\frac{1}{w})^2+2=3.$$ Similarly, we have $$w^4+\frac{1}{w^4}=(w^2+\frac{1}{w^2})^2-2=7$$ and $$w^8+\frac{1}{w^8}=47.$$ Since $$w^{10}+\frac{1}{w^{10}}=(w^2+\frac{1}{w^2})(w^8+\frac{1}{w^8})-(w^2+\frac{1}{w^2})(w^4+\frac{1}{w^4}-1),$$ we have $$w^{10}+\frac{1}{w^{10}}=3\times 47-3\times (7-1)=3\times 41 =123$$

alans
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  • Hi Alans. That's brilliant, but could you share how you managed to think of the 2nd last step? – Eric Aug 05 '16 at 15:31
  • @Eric I thought of $w^{10}+\frac{1}{w^{10}}$ as $(w^2)^5+(\frac{1}{w^{2}})^5$ and then used known factorisation $x^5+y^5=(x+y)(x^4-x^3 y+x^2 y^2-xy^3+y^4)$. – alans Aug 05 '16 at 15:56
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HINT:

If $F_n=w_1^n+w_2^n$ where $w_1,w_2$ are the roots of $w^2-w-1=0$

$F_0=1+1,F_1=w_1+w_2=1$

$F_{n+2}=F_{n+1}+F_n$

$n=0\implies F_{0+2}=F_{0+1}+F_0=2+1$

$F_{1+2}=F_{1+1}+F_1=3+2$

Now $w_1^{2m}+\dfrac1{w_1^{2m}}=w_1^{2m}+w_2^{2m}=?$

lab bhattacharjee
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