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Find $a$ such that $ax^{17}+bx^{16}+1$ is divisible by $x^2-x-1$.

I tried taking the roots of the polynomial which are $\frac{1±\sqrt{5}}{2}$

And I got the equation $a(\frac{1±\sqrt{5}}{2})^{17}+b(\frac{1±\sqrt{5}}{2})^{16}+1=0$

Now I don't know what to do next.

Any help would be appreciated.

Toby Mak
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Toshu
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  • Try $ax^3+bx^2+1$ first, remembering that $x^2=x+1$ so $x^3=x^2+x$. Then $ax^4+bx^3+1$ and try to spot a pattern. – Empy2 Oct 13 '19 at 06:30
  • Sorry, I can't spot any pattern. – Toshu Oct 13 '19 at 06:41
  • As in Robert's answer, they turn out to be Fibonacci numbers – Empy2 Oct 13 '19 at 06:43
  • @Empy2 Yes it looks like it has something to do with Fibonacci numbers but this question is there in the exercises in the end of the chapter and unfortunately there nothing given about Fibonacci numbers and I don't know anything about them. I believe there may be another way of solving this. – Toshu Oct 13 '19 at 06:50
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    It is just a sequence of numbers, but it crops up in many places. You will see it again. – Empy2 Oct 13 '19 at 07:09

4 Answers4

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Hint. If $x=\frac{1±\sqrt{5}}{2}$ then $x^2=x+1$ and \begin{align} ax^{17}+bx^{16}+1 &=(ax+b)(x+1)^8+1 \\ &=(ax+b)(x^2+2x+1)^4+1\\ &=(ax+b)(3x+2)^4+1\\ &=(ax+b)(21x+13)^2+1. \end{align} Can you take it from here? Note that 3,2, 21,13 are all Fibonacci numbers.

kelalaka
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Robert Z
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The given divisor has roots $\varphi=\frac{1+\sqrt5}2$ and $-\frac1\varphi$, so the $17$th-degree polynomial must also have these roots. This gives a system of equations for $a$ and $b$, which can then be solved.

The high exponents can be simplified by using the property $\varphi^2=\varphi+1$: $$\varphi^4=3\varphi+2$$ $$\varphi^8=21\varphi+13$$ $$\varphi^{16}=987\varphi+610$$ $$\varphi^{17}=1597\varphi+987$$ Thus $$a\varphi^{17}+b\varphi^{16}+1=0\implies a(1597\varphi+987)+b(987\varphi+610)=-1$$ $$-a\varphi^{-17}+b\varphi^{-16}+1=0\implies -a+b\varphi=-1597\varphi-987$$ We can see that $a=987,b=-1597$ is a solution to this system, and hence the original problem.

Parcly Taxel
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As you observed correctly, the 17th degree polynomial must also have the same roots ($\frac{1\pm\sqrt5}2$). Substituting these roots into $ax^{17}+bx^{16}+1=0$, you get two equations which can be solved simultaneously for $a$ and $b$, yielding $$a=987 \qquad \text{and} \qquad b=-1597.$$

Luke Collins
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Let $\alpha$ and $\beta$ be the two roots of the polynomial $x^2-x-1$. Observe that $\alpha\beta=-1$. Then, $$a\alpha^{17}+b\alpha^{16}+1=0$$ and $$a\beta^{17}+b\beta^{16}+1=0\,.$$ This means $$a\alpha+b=-\frac{1}{\alpha^{16}}=-\left(-\frac{1}{\alpha}\right)^{16}=-\beta^{16}\,.$$ Similarly, $$a\beta+b=-\alpha^{16}\,.$$ Consequently, $$(\alpha-\beta)\,a=\left(\alpha^{16}-\beta^{16}\right)\text{ or }a=\left(\frac{\alpha^{16}-\beta^{16}}{\alpha-\beta}\right)=F_{16}\,,$$ where $F_k$ is the $k$-th Fibonacci number (i.e., $F_0=0$, $F_1=1$, and $F_k=F_{k-1}+F_{k-2}$ for all integers $k$).

In the same manner, $$a-b\beta=a+\frac{b}{\alpha}=-\frac{1}{\alpha^{17}}=\beta^{17}$$ and $$a-b\alpha=a+\frac{b}{\beta}=-\frac{1}{\beta^{17}}=\alpha^{17}\,,$$ whence $$(\alpha-\beta)\,b=-\left(\alpha^{17}-\beta^{17}\right)\,.$$ Thus, $$b=-\left(\frac{\alpha^{17}-\beta^{17}}{\alpha-\beta}\right)=-F_{17}\,.$$

In general, if $n$ is a positive integer and $ax^{n+1}+bx^n+1\in\mathbb{R}[x]$ is divisible by $x^2-x-1$, then $$a=(-1)^n\,F_n\text{ and }b=(-1)^{n+1}\,F_{n+1}\,.$$ Even more generally, let $K$ be an arbitrary field, and $p$ and $q$ any elements of $K$ such that $q\neq 0$. Suppose that $ax^{n+1}+bx^n+1\in K[x]$ is divisible by $x^2+px+q$. Then, $$a=\frac{1}{q^n}\,f_n(p,q)\text{ and }b=-\frac{1}{q^n}\,f_{n+1}(p,q)\,,$$ where $f_0(p,q):=0$, $f_1(p,q):=1$, and $$f_k(p,q)+p\,f_{k-1}(p,q)+q\,f_{k-2}(p,q)=0$$ for all $k\in\mathbb{Z}$. The proof is very much the same (except for the case $p^2=4q$, which has to be dealt with separately).

Batominovski
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