B4 exercise: Express $$x=\sqrt[8]{2207-\cfrac{1}{2207-\cfrac{1}{2207-\cfrac{1}{2207-\cfrac{1}{\ddots}}}}}$$ as $\dfrac{a+b\sqrt{c}}{d}$, where $a,b,c,d$ are integers.
In the following solution https://kskedlaya.org/putnam-archive/1995s.pdf what is the idea behind computing
$$(x^2 - ax + 1)(x^2 + ax + 1) = x^4 - (a^2-2)x^2 + 1,$$
and why is the conclusion "the positive square roots of the quadratic $x^2 − bx + 1$ satisfy the quadratic $x^2 − x\sqrt{b^2 +2} + 1 = 0$"?
The inner expression $L$ satisfies the quadratic equation $L^2 - 2207L + 1 = 0 $.
The goal is to determine $ x = \sqrt[8]{L}$.
The approach we take is to find quadratic equations satisfied by $ \sqrt{L}$, then $\sqrt[4]{L}$, then $\sqrt[8]{L}$, which would allow us to calculate the value of $x$.
Claim: For $ a \geq 2$, suppose the roots of $ x^ 2 - ax + 1 = 0 $ are $ x_1, x_2 $ (which will be positive reals), then the roots of $ y^2 - \sqrt{a + 2 } y + 1 = 0 $ are $ \sqrt{ x_1}, \sqrt{ x_2} $.
My Proof: (Presenting this first as I believe it's easier to follow and hence understand why it is true.)
$\begin{align} & x_1 - \sqrt{ a + 2} \sqrt{ x_1} + 1 = 0 \\ \Leftrightarrow \quad & x_1 + 1 = \sqrt{a+2} \sqrt{x_1} \\ \Leftrightarrow \quad & x_1^2 + 2x_1 + 1 = (a+2)x_1 \\ \Leftrightarrow \quad & x_1^2 - ax_1 + 1 = 0 \\ \end{align}$
Note that the condition that $ a \geq 2$ gives us that the roots are positive reals, and so $ \sqrt{x_i}$ is unambiguous. A similar statement holds when $ a < 2$. (I encourage you to investigate this.) $_\square$
Kiran's proof: Using the transformation of roots $ \sqrt{x} = y \Leftrightarrow x = y^2$, we have $y^4 - ay^2 + 1 = 0 $.
The expression that OP asks about is $$(m^2 - \alpha m + 1 ) ( m^2 + \alpha m + 1) = m^4 - ( \alpha^2 - 2 ) m^2 + 1 $$ where I replaced the variables to reduce confusion within the context of proving this claim.
With this expression, letting $m = y, \alpha^2 - 2 = a$, we can see the factorization $$ (y^2 - \sqrt{a+2} y + 1) ( y^2 + \sqrt{a+2}y + 1) = y^4 - ay^2 + 1 = 0 .$$
Now, since the 2 square roots are positive, we have to reject $(y^2 + \sqrt{a+2} y + 1) = 0 $ (which has no positive roots), and hence they must satisfy $(y^2 - \sqrt{a+2} y + 1) = 0 $.
Note that the negative square roots satisfy $(y^2 + \sqrt{a+2} y + 1) = 0 $. $_\square$
Corollary:
Here is my take on the question:
$$x=\sqrt[8]{2207-\cfrac{1}{2207-\cfrac{1}{2207-\cfrac{1}{2207-\cfrac{1}{\ddots}}}}}$$
By raising everything to the power of $8$, we get:
$$x^8=2207-\cfrac{1}{2207-\cfrac{1}{2207-\cfrac{1}{2207-\cfrac{1}{\ddots}}}}$$
Manipulating the above expression, we get:
$$2207-x^8=\cfrac{1}{2207-\cfrac{1}{2207-\cfrac{1}{2207-\cfrac{1}{\ddots}}}}$$
Raising everything to the power of $-1$, we get:
$$\frac{1}{2207-x^8}=2207-\cfrac{1}{2207-\cfrac{1}{2207-\cfrac{1}{\ddots}}}$$
However, since the RHS of this expression is the same as the RHS of the expression after raising everything to the power of $8$:
$$\frac{1}{2207-x^8}=x^8$$
Manipulating the above, we are left with the following:
$$x^{16}-2207x^8-1=0$$
Letting $y=x^8$, we have: $$y^2-2207y+1=0$$
Using the quadratic formula to solve the above quadratic, we have:
$$y=\frac{2207 \pm \sqrt{2207^2-2^2}}{2}$$
Using the difference of two squares identity and prime factorisation:
$$2207^2-2^2=2205*2209=21^2*47^2*5$$
Therefore: $$y=\frac{2207 \pm 987\sqrt{5}}{2}$$
Calculating both potential values of $y$, we find that: $$\fbox{$y_1, y_2 > 0$}$$
But $y=x^8$.
Therefore: $$x=\sqrt[8]{\frac{2207 \pm 987\sqrt{5}}{2}}$$
However, the fact that if the first number of the continued fraction is $0$, then the number it represents is less than $1$ (and given that the first number of our continued fraction is $2207$), we find that:
$$\fbox{$x> 1$}$$
Calculating both potential values of $x$ and discarding the value that is below $1$, we find that:
$$x=\sqrt[8]{\frac{2207+987\sqrt{5}}{2}}$$
Now, we need to get $x$ to be in the form of $\frac{a+b\sqrt{c}}{d}$.
Therefore:
$$(a+b\sqrt{c})^8 = \frac{d^8}{2}(2207 + 987\sqrt{5})$$
Expanding the LHS and splitting the equation, we get the following:
$$a^8+28a^6b^2c+70a^4b^4c^2+28a^2b^6c^3+b^8c^4=\frac{2207d^8}{2}$$
$$(8a^7b+56a^5b^3c+56a^3b^5c^2+8ab^7c^3)\sqrt{c}=\frac{987d^8}{2}\sqrt{5}$$
From the second equation, we can deduce that:
$$\fbox{$c=5$}$$
Therefore, the equations now become:
$$a^8+140a^6b^2+1750a^4b^4+3500a^2b^6+625b^8=\frac{2207d^8}{2}$$
$$8a^7b+280a^5b^3+1400a^3b^5+1000ab^7=\frac{987d^8}{2}$$
This is where the solution ends.
Post-Solution Thoughts:
Since WolframAlpha says that:
$$\fbox{$\sqrt[8]{\frac{2207 + 987\sqrt{5}}{2}} \equiv \frac{3+\sqrt{5}}{2}$}$$
and given my approach to finding $x$, I was looking into different ways to find $a,b,d$. My attempts so far have included the following:
$$(a+b\sqrt{5})^8-(a-b\sqrt{5})^8=16ab\sqrt{5}(a^2+5b^2)(a^4+30a^2b^2+25b^4)=2207d^8$$
(this was found using the second-to-last set of equations and adding and subtracting to get $(a+b\sqrt{5})^8$ and $(a-b\sqrt{5})^8$ respectively then subtracting them and simplifying the expression on the LHS.)
I am unsure how the above helps at all. I have also tried adding the last two equations and seeing if the LHS of the resulting equation can simplify into the form of $(a+kb)^n$ but it doesn't look like that will be the case. The same applies for subtraction.
Given that I am struggling to see any methods to find $a,b,d$, I turn to the viewers of this solution to help me find any viable approaches that follow on from my solution to finding the rest of the values of $x$ .
Update to Post-Solution Thoughts: I have been recently thinking about $\frac{3+\sqrt{5}}{2}$ and noticed that $$\fbox{$d=a-b$}$$
However, I would need to think of some proof of it going from $d=a-b$ to an equation or result I have previously obtained. At the time of posting this update, it looked like a promising avenue of attack and would've certainly helped but after investigating it, it led me nowhere. Plus, the posting of Update 2 (based off a commenter's method that was looking a lot more promising (at the time of posting the first iteration of Update 2)) effectively ended the usability of this method.
Update 2 to Post-Solution Thoughts: Having received a comment talking about how $k^{\frac{1}{8}}$ is the same as doing $\sqrt{k}$ $3$ times, I decided to try it out.
First, we will do the following:
$$\sqrt{\frac{2207+987\sqrt{5}}{2}}=\frac{a_1+b_1\sqrt{c_1}}{d_1} \tag1$$ ($a_m, b_m, c_m, d_m$ are all constants to be figured out.)
Rearranging the above, we get
$$\frac{d_1^2}{2}(2207+987\sqrt{5})=(a_1+b_1\sqrt{c_1})^2 \tag2$$
Expanding and simplifying the RHS before breaking the terms into a set of equations, we get
$$a_1^2+c_1b_1^2=\frac{2207d_1^2}{2} \tag3$$
$$2a_1b_1\sqrt{c_1}=\frac{987d_1^2\sqrt{5}}{2} \tag4$$
$(4)$ implies that $$\fbox{$c_1=5$}$$
This therefore means that $(3)$, $(4)$ becomes
$$a_1^2+5b_1^2=\frac{2207d_1^2}{2} \tag5$$
$$2a_1b_1=\frac{987d_1^2}{2} \tag6$$
Solving for $d_1^2$ then merging the $2$ equations, we get
$$a_1b_1=\frac{987(2a_1^2+10b_1^2)}{4(2207)} \tag7$$
Rearranging and simplifying, we get
$$0=987a_1^2-4414a_1b_1+4935b_1^2 \tag8$$
Letting the roots of the equation be $k_1,k_2$, we find (using Viete's formulae for the roots of a polynomial) that
$$k_1+k_2=\frac{4414}{987} \tag9$$
$$k_1k_2=5 \tag{10}$$
Rearranging the second equation and substituting into the first equation gives us
$$\frac{5}{k_f}+k_f=\frac{4414}{987} \tag{11}$$
(where $f$ can be $1$ or $2$)
Manipulating, rearranging and solving $(11)$ gives us the following pairs of solutions to $(8)$:
$(k_1,k_2)$ $=$ $(\frac{105}{47}$ , $\frac{47}{21})$ , $(\frac{47}{21}$ , $\frac{105}{47})$
A comment explained that my roots are equivalent to $\frac{a_1}{b_1}$.
Therefore, the above roots now become (after rearranging)
$$\fbox{$b_1=\frac{21a_1}{47}$, $b_1=\frac{47a_1}{105}$}$$
Now, substituting these roots back into $(6)$ and solving for $d_1$ , we get the following pairs of solutions to $(5)$ , $(6)$
$$\fbox{$b_1=\frac{21a_1}{47}$, $d_1=\pm\frac{2a_1}{47}$}$$
$$\fbox{$b_1=\frac{47a_1}{105}$, $d_1=\pm\frac{2a_1}{21\sqrt{5}}$}$$
Whilst investigating ways to find $a_1, b_1, d_1$, it occurred to me that if we proved that
$$\sqrt{\frac{e_1 + f_1\sqrt{g_1}}{h_1}} \equiv \frac{i_1+j_1\sqrt{l_1}}{m_1} \tag{12}$$
and that $e_1, f_1, g_1, h_1, i_1, j_1, l_1, m_1$ are all positive integers, then the first root pair would be the only valid root pair since in the second root pair, the denominator of $d_1$ is irrational. This would also mean that $47 \vert a_1$ and $2\vert d_1$.
Proof of $(12)$:
$$\sqrt{\frac{e_1 + f_1\sqrt{g_1}}{h_1}} = \frac{i_1+j_1\sqrt{l_1}}{m_1}$$
Rearranging, expanding and simplifying $(12)$, we get:
$$\frac{m_1^2}{h_1}(e_1 + f_1\sqrt{g_1})=i_1^2+2i_1j_1\sqrt{l_1}+g_1j_1^2$$
Splitting the equation into a pair of simultaneous equations:
$$\frac{m_1^2}{h_1}e_1=i_1^2+g_1j_1^2$$
$$\frac{m_1^2}{h_1}f_1\sqrt{g_1}=2i_1j_1\sqrt{l_1}$$
Splitting the second equation into another pair of simultaneous equations:
$$\frac{m_1^2}{h_1}f_1=2i_1j_1$$
$$\sqrt{g_1}=\sqrt{l_1}$$
Solving the second equation, we get:
$$\fbox{$g_1=l_1$}$$
Therefore, the original pair of simultaneous equations becomes:
$$\frac{m_1^2}{h_1}e_1=i_1^2+g_1j_1^2$$
$$\frac{m_1^2}{h_1}f_1=2i_1j_1$$
Manipulating the above equations such that the RHS of each equation is equal to $\frac{m_1^2}{h_1}$ , we find that
$$\frac{2i_1j_1}{f_1}=\frac{i_1^2+g_1j_1^2}{e_1}$$
Rearranging the above, we find that it is equivalent to
$$0=f_1i_1^2-2e_1i_1j_1+f_1g_1j_1^2$$
Letting the solutions of the above polynomial be $A_1, A_2$ , and using Viete's roots of polynomials formulae for a quadratic, we find that
$$A_1 + A_2=\frac{2e_1}{f_1}$$ $$A_1A_2=g_1$$
Manipulating, rearranging and solving the first equation (using the second equation as a substitution), we get
$$A_C = \frac{e_1}{f_1} \pm \sqrt{\frac{e_1^2}{f_1^2}-g_1}$$
(where $C$ can equal $1$ or $2$ .)
Therefore, the solution to the set of Viete's polynomial root equations (and hence, the original polynomial) are
$(A_1, A_2) = (\frac{e_1}{f_1} + \sqrt{\frac{e_1^2}{f_1^2}-g_1}, \frac{e_1}{f_1} - \sqrt{\frac{e_1^2}{f_1^2}-g_1}), (\frac{e_1}{f_1} - \sqrt{\frac{e_1^2}{f_1^2}-g_1}, \frac{e_1}{f_1} + \sqrt{\frac{e_1^2}{f_1^2}-g_1})$
The roots given above are of the form $\frac{i_1}{j_1}$ .
Rearranging the roots, we get
$$\fbox{$i_1 = j_1^2(\frac{e_1}{f_1} + \sqrt{\frac{e_1^2}{f_1^2}-g_1}), j_1^2(\frac{e_1}{f_1} - \sqrt{\frac{e_1^2}{f_1^2}-g_1})$}$$
Putting both solutions into the original set of simultaneous equations derived from $(12)$ , we get the following pair of solutions
$$\fbox{$i_1 = j_1^2(\frac{e_1}{f_1} + \sqrt{\frac{e_1^2}{f_1^2}-g_1}), j_1 = \pm \frac{m\sqrt[4]{(e_1f_1)^2+ g_1)}}{\sqrt{2h(e_1+f_1)\sqrt{\frac{e_1^4}{f_1^4}-g_1^2}}}$}$$
$$\fbox{$i_1 = j_1^2(\frac{e_1}{f_1} - \sqrt{\frac{e_1^2}{f_1^2}-g_1}), j_1 = \pm \frac{m\sqrt[4]{(e_1f_1)^2+ g_1)}}{\sqrt{2h(e_1-f_1)\sqrt{\frac{e_1^4}{f_1^4}-g_1^2}}}$}$$
(Rest of proof TBD at a later date)
Also, noticed the following: if you square both values of $d_1$ from both root pairs and compare the denominator, you get
$$d_1^2=\pm\frac{4a_1^2}{2207\pm2} \tag{13}$$
This means that $(13)$ can be used to rewrite the pairs of roots as the following
$$d_1^2=\pm \begin{cases} \frac{4a_1^2}{2207+2}, & \text{if $b_1=\frac{21a_1}{47}$} \\ \frac{4a_1^2}{2207-2}, & \text{if $b_1=\frac{47a_1}{105}$} \end{cases}$$
$(13)$ also shows that the denominator is of the form $a\pm2$, which indicates some connection to part of the original question the OP was asking - will need to investigate this further.
Edit 1: Despite this question now being labeled as a duplicate, I will still continue to edit and refine this solution until it is complete (i.e. I have solved the equation and have answered the OP's original question.).
Any suggestions for/edits to this solution are welcome.
Since $x^8=2207-\frac1{x^8}$, $$x^{16}-2207x^8+1=0$$ $$(x^2-3x+1)(x^2+3x+1)(x^4+7x^2+1)(x^8+47x^4+2)=0.$$ The real roots are coming from the first factor.
$$x^2-3x+1=0\implies x=\frac{3+\sqrt5}2$$
since the other root is less than $1$.
