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Problem:

Find $(a,b)$ such that $$L_n = a\phi^n + b\widehat{\phi}^n.$$

Where $n$ is the $n^{th}$ lucas number.

How would I start this? Would I just start by plugging in $a=b=1$ and then trying to solve?

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  • $\begingroup$ You can do this by noting that the generating function of the sequence $a_{n+2} = a_{n+1} + a_n$ is $\frac{a_0-a_0x+a_1x}{1-x-x^2}$ and using partial fractions. $\endgroup$ – Patrick Stevens Apr 2 '17 at 13:35
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$$L_0=2$$

$$L_1=1$$

Thus,

$$a+b=2$$

$$a\phi+b\hat\phi=1$$

And so,

$$a=2-b$$

$$(2-b)\phi+b\hat\phi=1$$

$$\implies b=\frac{1-2\phi}{\hat\phi-\phi}$$

$$\implies a=2-\frac{1-2\phi}{\hat\phi-\phi}$$

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  • $\begingroup$ I'm having a difficult time with this question. It's well known that $L_n = \phi^n + \widehat{\phi}^n$. So why don't you just finish what you started and say that $a=b=1$? $\endgroup$ – Cye Waldman Apr 4 '17 at 22:37
  • $\begingroup$ I didn't finish because I've already taken care of the bulk of the work for you. All you really need to do now is substitute $\phi=\frac{1+\sqrt5}2$ and $\hat\phi=\frac{1-\sqrt5}2$. $\endgroup$ – Simply Beautiful Art Apr 4 '17 at 22:40
  • $\begingroup$ My problem is not with you or your solution, it's the fact it legitimizes the original question. The problem being that if $a$ and $b$ are not unity then you could not have a solution in terms of $\phi$ in the first place. See my posting in math.stackexchange.com/questions/2152925/… $\endgroup$ – Cye Waldman Apr 5 '17 at 0:51

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