MHB Solve x^2-x-1: Find a & b | Integer Solution

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The discussion revolves around finding integer values for 'a' and 'b' such that the polynomial ax^17 + bx^16 + 1 is divisible by x^2 - x - 1. The roots of the polynomial x^2 - x - 1 are identified, and the remainder theorem is applied to set up equations based on these roots. The relationship between the coefficients and Fibonacci numbers is established, leading to the equation aF_17 + bF_16 = 0. An alternative method is suggested, involving the multiplication of x^2 - x + 1 with another polynomial to derive coefficients for the higher powers of x. The discussion highlights the connection between the coefficients and Fibonacci sequences, indicating a pattern that needs to be explored further.
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Find integers ‘a’ and ‘b’ such that x2-x-1 divides ax17+bx16+1=0

Please help
 
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I would first consider that the roots of:

$$x^2-x-1$$

are:

$$x=\frac{1\pm\sqrt{5}}{2}$$

and so by the remainder theorem, we want:

$$f\left(\frac{1+\sqrt{5}}{2} \right)=a\left(\frac{1+\sqrt{5}}{2} \right)^{17}+b\left(\frac{1+\sqrt{5}}{2} \right)^{16}+1=0$$

$$f\left(\frac{1-\sqrt{5}}{2} \right)=a\left(\frac{1-\sqrt{5}}{2} \right)^{17}+b\left(\frac{1-\sqrt{5}}{2} \right)^{16}+1=0$$

If we subtract the second equation from the first, we obtain:

$$a\left(\left(\frac{1+\sqrt{5}}{2} \right)^{17}-\left(\frac{1-\sqrt{5}}{2} \right)^{17} \right)+b\left(\left(\frac{1+\sqrt{5}}{2} \right)^{16}-\left(\frac{1-\sqrt{5}}{2} \right)^{16} \right)=0$$

If we divide through by $$\sqrt{5}$$ we may write:

$$a\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2} \right)^{17}-\left(\frac{1-\sqrt{5}}{2} \right)^{17} \right)+b\frac{1}{\sqrt{5}}\left(\left(\frac{1+ \sqrt{5}}{2} \right)^{16}-\left(\frac{1-\sqrt{5}}{2} \right)^{16} \right)=0$$

Using the fact that the closed form for the $n$th Fibonacci number is:

$$F_n=\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2} \right)^{n}-\left(\frac{1-\sqrt{5}}{2} \right)^{n} \right)$$

we may now write:

$$aF_{17}+bF_{16}=0$$

Can you finish?
 
suvadip said:
Find integers ‘a’ and ‘b’ such that x2-x-1 divides ax17+bx16+1=0
Here is another method. Form a product of $1+x-x^2$ with another polynomial in such a way that, after the initial constant term $1$, each coefficient in the product is $0$ until you reach the powers $x^{16}$ and $x^{17}.$ The calculation will have to start like this: $$(1+x-x^2)(1-x +\ldots)$$ (the coefficient of $x$ in the second bracket must be $-1$ in order for the $x$-term in the product to have coefficient $0$). As you continue to feed in further powers of $x$, you will find that the product will look like this: $$(1+x-x^2)(1-x + 2x^2 - 3x^3 + 5x^4 - 8x^5 + \ldots).$$ What do you notice about the sequence of coefficients, and can you continue the pattern?[I think we had this same problem somewhere in this forum a few months ago, but I can't find it.]
 
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