Solve for all $x$ such that $x^3 = 2x + 1, x^4 = 3x + 2, x^5 = 5x + 3, x^6 = 8x +5 \cdots$

Question

Question:

Solve for all $x$ such that $\begin{cases}&{x}^{3}=2x+1\\&{x}^{4}=3x+2\\&{x}^{5}=5x+3\\&{x}^{6}=8x+5\\&\vdots\end{cases}$.

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My attempt:

I sum up everything. $$\begin{aligned}\sum_{i=1}^n x^{i+2} &= (2 + 3 + 5 + 8 + \cdots)x + (1 + 2 + 3 + 5 +...)\\&= (S_n - 2)x + (S_n - 1)\\& = (a_{n+2} - 3)x + (a_{n+2} - 2)\end{aligned}$$

where $S_n$ is sum of first $n$ terms of Fibonacci series and $a_n$ is its $n$th term.

I'm not getting any idea how to solve it further.

Answer 1

Hint:

As $x(2x+1)\ne0,$

$$\dfrac{x^4}{x^3}=\dfrac{3x+2}{2x+1}\iff x^2-x-1=0$$

Similarly check, $$\dfrac{x^5}{x^4}=?,\dfrac{x^6}{x^5}=?$$

Find the common root of all?

Answer 2

$x^3 = 2x + 1$ has an evident rational solution, that being $-1.$ Next, $x+1$ divides $x^3 - 2x - 1,$ the quotient is $x^2 - x-1$ for $$ x^3 - 2x - 1 = (x+1) (x^2 - x - 1) $$ However $-1$ does not satisfy $x^4 = 3x + 2.$

We are left with those numbers $x$ with $x^2 = x+1.$ We can find the values of $x^n$ by a simple rule: if $x^n = ax + b,$ then $$x^{n+1} = a x^2 + bx = a (x+1) + bx = (a+b)x + a. $$ The coefficient pairs for $x^n$ come out $$ \begin{array}{ccc} 2: & 1 & 1 \\ 3: & 2 & 1 \\ 4 : & 3 & 2 \\ 5 : & 5 & 3 \\ \end{array} $$ and so on. Given a row $(a,b) $ the next row is what happens when we multiply it on the right side by $$ M= \left( \begin{array}{cc} 1 & 1 \\ 1 & 0 \\ \end{array} \right) $$ All of the pairs are of the form $(1 \hspace{2mm} 1) \; M^j . $ Actually, now that I think of it, all of the pairs are of the form $(1 \hspace{2mm} 0) \; M^k $ and are just the top row of that $M^k. \;$ These are, of course, consecutive fibonacci numbers

Answer 3

$$x^4=x^3x=2x^2+x\implies3x+2=2x^2+x\implies-2x^2+2x+2=0\implies x^2-x-1=0$$ So $x=\frac{1\pm\sqrt{5}}{2}=\varphi,\varphi^{\dagger}$. You could use induction to prove that they are solutions to all other equations.