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\(\displaystyle x^2-x-1\)

are:

\(\displaystyle x=\frac{1\pm\sqrt{5}}{2}\)

and so by the remainder theorem, we want:

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

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

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

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

If we divide through by \(\displaystyle \sqrt{5}\) we may write:

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

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

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

we may now write:

\(\displaystyle aF_{17}+bF_{16}=0\)

Can you finish?

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- Feb 7, 2012

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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?Find integers ‘a’ and ‘b’ such that x^{2}-x-1 divides ax^{17}+bx^{16}+1=0

[I think we had this same problem somewhere in this forum a few months ago, but I can't find it.]