# If $a^2-a-1=0$ where $a\gt0$, then what does $a^6$ equal? (Olympiad question)

$$\color{white}{\require{cancel}{3}}$$

So I was looking on Youtube for math equations that I thought that I could probably solve when I came across this video by the channel Maths and many more. The question in this video was$$\text{If }a^2-a-1=0\text{, then what is the value of }a^6\text{, where }a\gt0$$which I wanted to try and solve on my own. Here are the steps that I took to solve the equation:$$\text{Put it into the quadratic formula since it is unfactorable with rational numbers}$$$$\frac{1\pm\sqrt{1-4(1)(-1)}}{2}$$$$=\frac{1\pm\sqrt{5}}{2}$$however, since the value of $$a$$ is greater than $$0$$, it is actually $$a=\frac{1+\sqrt{5}}{2}$$$$\text{Then, all you have to do is square it}$$$$\left(\frac{1+\sqrt{5}}{2}\right)^2$$$$\implies\frac{1}{4}(1+\sqrt{5})^2\text{, and simplifying that gets}$$$$\frac{1}{4}(6+2\sqrt{5})$$$$\iff\frac{6+2\sqrt{5}}{4}$$$$\iff1.5+\frac{\sqrt{5}}{2}=a^2$$$$\implies(1.5+\frac{\sqrt{5}}{2})^2=a^4$$

$$\color{white}{.}$$ $$1.5$$ $$\frac{\sqrt{5}}{2}$$
$$1.5$$ $$2.25$$ $$\frac{3\sqrt{5}}{4}$$
$$\frac{\sqrt{5}}{2}$$ $$\frac{3\sqrt{5}}{4}$$ $$1.25$$

$$2.25+2\left(\frac{3\sqrt{5}}{4}\right)+1.25$$$$\iff3.5+\frac{3\sqrt{5}}{2}=a^4$$

$$\color{white}{.}$$ $$3.5$$ $$\frac{3\sqrt{5}}{2}$$
$$1.5$$ $$5.25$$ $$2.25\sqrt{5}$$
$$\frac{\sqrt{5}}{2}$$ $$1.75\sqrt{5}$$ $$3.75$$

$$5.25+(2.25+1.75)\sqrt{5}+3.75=a^6$$$$\implies\therefore9+4\sqrt{5}=a^6$$$$\text{And,when we plug it into Wolfram Alpha:}$$ $$\text{We get the same thing! :)}$$

My question

Is my solution correct, and if not, what could I do to attain the correct solution or what could I do to attain it more easily?

To clarify

1. Sorry if this seems like a trivial/short question
2. If you want to edit this question to improve it, I am so sorry about the compressed formatting, it's just easier for me to type it like this for it to be faster.
3. Sorry if the tags aren't correct, they most likely are but still.
4. Sorry if I used any math functions incorrectly.
• Also sorry about all of the edits right after I posted it, I found out that my tables sort of broke :\ Apr 25 at 14:56
• Your solution is fine. I think you can get a slightly quicker result by using $a^2 = a+1$ to deduce that $a^4 = 3a+2$, but I doubt it can be done without the quadratic formula. So it's not much different from your approach. Apr 25 at 14:59
• @preferred_anon Is that because $a^2-a-1$ can't really be factored using the factoring methods that most people would know? Apr 25 at 15:00
• Just use $a^2=a+1$ to decrease the degree. Apr 25 at 15:02

Your solution is fine, but there are easier ways to do this problem. Specifically, notice that

$$a^2=a+1$$

implies

$$a^6=(a^2)^3=(a+1)^3=a^3+3a^2+3a+1$$ $$=a(a+1)+3(a+1)+3a+1=a^2+a+6a+4$$

$$=(a+1)+7a+4=8a+5$$

Then using the $$a$$ you found the answer is

$$a^6=4+4\sqrt{5}+5=9+4\sqrt{5}$$

$$a^2=a+1$$ Multiply by $$a$$ four times, subsituting the above expression for $$a^2$$ at each stage:$$a^3=a^2\cdot a=a^2+a=2a+1$$ $$a^4=a^3\cdot a=2a^2+a=3a+2$$ $$a^5=a^4\cdot a=3a^2+2a=5a+3$$ $$a^6=a^5\cdot a=5a^2+3a=8a+5$$

Now plug in your solution $$a=\dfrac{1+\sqrt 5}{2}$$.

• Thanks but what would I do after getting that $a^6$ equals $8a+5$? (if I was going off of your solution) Would I have the previous knowledge to know that $a=\frac{1+\sqrt{5}}{2}$? Please explain. Apr 25 at 15:21
• Yes, I was assuming that you had the solution to $a^2-a-1=0$. I edited my answer accordingly. Apr 25 at 15:34
• @TonyK (+1) Would you like to find the closed form for $a^n~?$ $~~\ddot\smile$ Apr 25 at 15:37
• Note that once you have $a^3$ you can prove $a^6=8a+5$ by squaring, but then the pattern @lonestudent hints at is less obvious.
– J.G.
Apr 25 at 15:44
• @lonestudent Would the closed form for $a^n$ be $a^n=a^{n-1}+a^{n-2}$? Apr 25 at 16:20

Let $$\thinspace a=\frac {1+\sqrt 5}{2}$$, then using high school algebra, you have :

\begin{align}&a^2-a+1=2\\ \implies &a^3+1=2(a+1)\\ \implies &a^3=2a+1\\ \implies &a^6=4(a+1)+4a+1\\ &\,\,\,\,\;=8a+5\,.\end{align}

• Where did the 2 come from? I'm confused. Apr 25 at 17:28
• @CrSb0001 A well-known formula: \begin{align}a^3+b^3&=(a+b)(a^2-ab+b^2)\end{align} Apr 25 at 17:40

The Euclidean (long) division of polynomial $$\,a^6\,$$ by $$\,a^2 - a - 1\,$$ gives $$\,a^6\,$$ in terms of $$\,a\,$$ directly:

$$\require{cancel} a^6 = \cancel{(a^4 + a^3 + 2 a^2 + 3 a + 5)(a^2 - a - 1)} + (8 a + 5)$$

Since $$a={1\over 2}(1+\sqrt{5})$$ we get $$a^n=c_n+d_n\sqrt{5}$$ for some positive rational numbers $$c_n$$ and $$d_n.$$ We have $$c_1=d_1={1\over 2}.$$ The condition $$a^2=1+a$$ gives $$c_2=c_1+1={3\over 2},\ d_2=d_1={1\over 2}$$. In view of $$a^{n+2}=a^n+a^{n+1}$$ we get $$c_{n+2}=c_n+c_{n+1},\quad d_{n+2}=d_n+d_{n+1},\quad n\ge 1$$ Thus $$c_6=c_4+c_5=c_3+2c_4=2c_2+3c_3=3c_1+5c_2=9$$ Similarly $$d_6=3d_1+5d_2=4.$$

We can continue that way in order to determine higher powers of $$a.$$

We also have

$$a^2 \ - \ a \ - \ 1 \ \ = \ \ 0 \ \ \Rightarrow \ \ a^2 \ - \ 1 \ \ = \ \ a \ \ \Rightarrow \ \ a \ - \ \frac{1}{a} \ \ = \ \ 1$$ $$\Rightarrow \ \ \left( \ a \ - \ \frac{1}{a} \ \right)^3 \ \ = \ \ a^3 \ - \ 3a \ + \ \frac{3}{a} \ - \ \frac{1}{a^3} \ \ = \ \ a^3 \ - \ 3·\left(a \ - \ \frac{1}{a} \right) \ - \ \frac{1}{a^3} \ \ = \ \ 1$$ $$\Rightarrow \ \ a^3 \ - \ \frac{1}{a^3} \ \ = \ \ 4$$ $$\Rightarrow \ \ \left( \ a^3 \ - \ \frac{1}{a^3} \ \right)^2 \ \ = \ \ a^6 \ - \ 2 \ + \ \frac{1}{a^6} \ \ = \ \ 16 \ \ \Rightarrow \ \ a^6 \ + \ \frac{1}{a^6} \ \ = \ \ 18 \ \ .$$

If we now treat this as a quadratic equation in $$\ u \ = \ a^6 \ \ ,$$ we conclude that $$u \ + \ \frac{1}{u} \ = \ 18 \ \ \Rightarrow \ \ u^2 \ - \ 18u \ + \ 1 \ \ = \ \ 0$$ $$\Rightarrow \ \ u \ \ = \ \ a^6 \ \ = \ \ \frac{18 \ \pm \ \sqrt{18^2 \ - \ 4}}{2} \ \ = \ \ 9 \ \pm \ \frac{\sqrt{320}}{2} \ \ = \ \ 9 \ \pm \ \sqrt{80} \ \ = \ \ 9 \ \pm \ 4\sqrt5 \ \ ,$$ the two values corresponding to $$\ a \ = \ \frac{1 \ \pm \sqrt5}{2} \ \$$ (which we did not need to know in advance). So the answer for the problem is $$\ a^6 \ = \ 9 + 4\sqrt5 \ \ .$$

The Fibonacci numbers are $$F_{-1}=1,\,F_{0}=0,\, F_1=1,\, F_2=1,\, F_3=2,\, F_4=3,\, F_5=5,\, F_6=8,\, F_7=13, ...$$ They are defined by the reccurrence relation $$F_{k+1}=F_k+F_{k-1}$$.

Claim: $$a^{k}=F_{k}a+F_{k-1}$$ for all $$k\in\Bbb N.$$

Proof: (By induction) $$a^0=1=F_0a+F_{-1}=0.a+1=1$$ is true. Assume that $$a^{i}=F_{i}a+F_{i-1}$$ for all $$0\leq i\leq k$$. Then

$$a^{k+1}=a^ka=(F_ka+F_{k-1})a=F_ka^2+F_{k-1}a=F_ka+F_k+F_{k-1}a=F_{k+1}a+F_k$$ completing the induction.

Here, $$a=\phi=\frac{\sqrt 5+1}{2}$$ is the golden-ratio and it is the positive root of $$a^2-a-1=0$$.
$$\phi\approx 1.6180339887498948482045868343656381177203091798057628621354486227$$