Golden Indeed

By experimenting with the first few powers of $\varphi$ we get:

$\varphi=0+(1)\varphi$

$\varphi^2=1+(1)\varphi$

$\varphi^3=1+2\varphi$

$\varphi^4=2+3\varphi$

$\varphi^5=3+5\varphi$

$\varphi^6=5+8\varphi$

From here we see a relation to the Fibonnaci sequence. We can conjure:

$\varphi^n=F(n-1)+F(n)\varphi$

Where $F(n)$ is the fibonnaci sequence with $F(0)=1, F(1)+1 and F(n+1)=F(n)+F(n-1)$ for integers $n$ .

We can prove it by induction. The iductive step would be:

$\varphi^{n+1}=\varphi\varphi^n$

$=\varphi(F(n-1)+F(n)\varphi)$

$=F(n-1)\varphi+F(n)\varphi^2$

$=F(n-1)\varphi+F(n)(1+\varphi)$

$=F(n)+(F(n-1)+F(n))\varphi$

$=F(n)+F(n+1)\varphi$

Hence $\varphi^{14}=F(13)+F(14)\varphi$ . So we need $F(13)+F(14)=F(15)=610$

Consider the fact that $\phi$ is the solution to the quadratic $\phi^2 - \phi - 1 = 0$ , then it better be the case that $\phi^2 = 1 + \phi$ . By elimination, it's easy to see that the extension $\mathbb{Z}[\phi]$ (all numbers of the form $a + b\phi$ ) is closed under addition and multiplication! $\begin{aligned}(a+b\phi)(c+d\phi) &= ac + (bc +ad)\phi + bd\phi^2 \\ &= (ac + bd) + (bc + ad + bd)\phi\end{aligned}$ therefore arithmetic on $\mathbb{Z}[\phi]$ can be alternatively characterized on the field $\mathbb{\Phi} = (\mathbb{Z}^2, \oplus, \otimes)$ whereby $\oplus$ is just addition over integer componentwise and $(a,b) \otimes (c,d) = (ac + bd, ad + bc + bd)$ Now, recall that $(a+b\phi)\phi = a\phi + b\phi^2 = b + (a+b)\phi$ , so $(a,b)(0,1) = (b,a+b)$ : $\begin{aligned}\phi &= (0,1) \\ \phi^2 &= (0,1)(0,1) = (1,1) \\ \phi^3 &= (1,1)(0,1) = (1,2) \\ \phi^4 &= (2,3) \\ \phi^5 &= (3,5) \\ \vdots\end{aligned}$ let the second component of $\phi^k$ be $f_k$ and the first component $f_{k-1}$ , then $(f_k,f_{k+1}) = (f_{k-1},f_k)(0,1) = (f_k,f_{k-1}+f_k)$ , therefore, this is entirely characterized by the recurrence $f_0 = 0, f_1 = 1, f_{k+1} = f_k + f_{k-1}$ , which turns out to be the fibonacci numbers! Anyways, this gives $\phi^k = (f_{k-1},f_k)$ so the value of $a+b$ for $\phi^{14}$ must be $f_{13} + f_{14} = f_{15}$ , which in this case is $\boxed{610}$

It is known that $\phi ^ n = F(n) \times \phi + F(n-1)$ , where $F(n)$ is the Fibonacci's n-th number.

By the Fibonacci Sequence recursion definition, $F(n-1) + F(n) = F(n+1)$ . This means $a+b$ will be the 15-th Fibonacci number, which happens to be $610$ .

$\varphi$ satisfy the quadratic equation $\varphi^2-\varphi-1=0$ . So we have $\varphi^3=\varphi^2+\varphi=2\varphi+1$ And $\varphi^4=\varphi^2+2\varphi+1=3\varphi+2.$ So we can calculate that $\varphi^7=6\varphi^2+7\varphi+2=13\varphi+8$ And we have the result $\varphi^{14}=169\varphi^2+208\varphi+64=377\varphi+233$ With $a=233$ and $b=377$ , we have $a+b=610$

We prove by induction $\varphi ^n=f_n\varphi+f_{n-1}$ We have $\varphi ^2=\varphi+1$ Assume that the statement holds for n. $\varphi^{n+1}=\varphi (f_n\varphi+f_{n-1})=f_n(\varphi +1)+f_{n-1}\varphi=f_{n+1}\varphi+f_n$ Then the statement also holds for n+1. So $a+b=f_{14}+f_{13}=f_{15}=610$

ø^14=233+377ø

233+377=610

Powers of the Golden Ratio satisfy the same recurrence relation as the Fibonacci numbers, namely $a_{n} = a_{n-1}+a_{n-2}$ . From this, we see that that $\phi^{n} = F_{n}*\phi + F_{n-1}$ , where $F_{n}$ denotes the $n$ th Fibonacci number. Therefore, $\phi^{14}$ is equivalent to $F_{14}+F_{13} = F_{15} = \boxed{610}$

First you must notice that given the formula $\phi^{n} = \phi^{n-1} + \phi^{n-2}$ it yields that for $n \geq 2$ , $\phi^{n} = F_{n-2} \phi^{2} + F_{n-3} \phi$ Where $F_{i}$ is the i-th Fibonacci number. Then it is easy to see that $\phi^{14} = 367 \phi + 233$ . Therefore $a + b = 233 + 367 = 610$ .

Given x as a golden ratio, we have the equation x^2 = x + 1 Therefore, x^2 - x = 1 Therefore, we can prove that x^n = x^(n-1) + x(n-2) Because x^n - x^(n-1) = x^(n-2) [x^2 - x] = x^(n-2) So the a and b ratio will follow the Fibonanci series a(1) = a(2) = 1; a(n) = a(n-1) + a(n-2) b(1) =0; b(2) = 1; b(n) = b(n-1) + b(n-2)