The first three quarters of me working on this problem were to find a distinctly unique pattern that works for most values of $(a,b,c)$. As I will elaborate in the rest of the Motivation, such patterns are not easy to trace (however, in the current 2019-2020 meta where Farey sequences and the expression $a_ib_j - a_jb_i = 1$ appeared again in abundance: examples include RMM Problem 2, AMO Problem 4 and GQMO problem 3 --- I'm not so sure ... perhaps those problems subconsciously altered my initial train of thought when I tried to find patterns.) Since when does a Vieta Jumping on multiple variables can appear in an A2? \[\begin{tabular}{p{12cm}} $\rule{4cm}{0.5pt} \hspace{1cm} \bigstar \bigstar \hspace{1cm} \rule{4cm}{0.5pt}$ \\ \end{tabular} \\ \]Before proceeding further, here is the glossary of beautiful pairs which motivated me towards the Solution --- it would be a great exercise for the reader to try to find $``\text{quantum ties}"$ along different triples --- although the last two pairs have a clear resemblance to each other. \[ \begin{tabular}{c|c} $(a,b,c)$&$\{x_i\}, \{y_i\}$ \\ \hline $(2,4,2)$&$\{1,1\}, \{2,0\}$ \\ \hline $(2,4,0)$&$\{1,1,0\}, \{0,0,2\}$ \\ \hline $(2,3,2)$&$\{1,1,0\}, \{1,1,1\}$ \\ \hline $(2,5,3)$&$\{1,1\}, \{2,1\}$ \\ \hline $(5,13,8)$&$\{2,1\}, \{3,2\}$ \\ \hline $(3,41,11)$&$\{1,1,1\}, \{4,4,3\}$ \\ \hline $(3,41,10)$&$\{1,1,1,0,0\}, \{4,4,2,2,1\}$ \\ \hline $(7,10,8)$&$\{2,1,1,1\}, \{2,2,1,1\}$ \\ \end{tabular} \quad \Rightarrow \quad \begin{tabular}{c|c} $(a,b,c)$&$\{x_i\}, \{y_i\}$ \\ \hline $(8,11,9)$&$8$ ones / $7$ ones, $1$ two \\ \hline $(7,10,8)$&$7$ ones / $6$ ones, $1$ two \\ \hline $(8,13,10)$&$\{2,2\}, \{3,2\}$ \\ \hline $(5,6,4)$& A million ways; use zeroes! \\ \hline $(4,10,6)$&$\{1,1,1,1\}, \{2,2,1,1\}$ \\ \hline $(4,13,9)$&$\{1,1,1,1\}, \{2,2,2,1\}$ \\ \hline $(9,22,14)$&$\{2,2,1\}, \{3,3,2\}$ \\ \hline $(9,3,5)$&$\{2,2,1\}, \{1,1,1\}$ \\ \end{tabular} \]Here's a brief description of the tabulars above: I did not explicitly consider the cases where $ab=c^2$ (though, of course, it will add to the richness of the configurations later) because the entries of the sequences $\{x_i\}$ and $\{y_i\}$ would be boring --- it would require basically $\{x_i\}$ linearly proportionate to $\{y_i\}$, and I wanted a telling clue which works for all bounds $ab\geq c^2$, both weak and strong. The first three tabular entries are just a casual observation: what would happen if I were to set $c$ equal to the smaller number between $a$ and $b$? (which, for almost the rest of the tabular entries, I'd assume to be $a$, with the exception of the last algorithmically correct one) Adjusting $c$ to be zero and $b$ to be tighter yielded more interesting configurations, as when $c$ is forced to be zero then all pairs $(x_i,y_i)$ for $1 \leq i \leq n$ must contain a zero. Furthermore, as $ab$ is slided closer to $c^2$, the sequences must be at close proximity with each other: at $(a,b,c) = (2,4,2)$ the terms consist of the loosely related $\{1,1\}$ with $\{2,0\}$. Compare that with the $\{1,1,0\}$ and $\{1,1,1\}$ which must occur when $(a,b,c) = (2,3,2)$. Now, aside from the weaker bounds I stretched upon in the beginning, I began to shift toward the tighter bound: what are the possible pairs when $ab = c^2+1$? It turns out that $n$ apparently must be $2$, with terms similar to Fibonacci numbers. Here the thought of Vieta Jumping has already sprouted, but here it's not immediately clear what to base it on: $a$, $b$, or $c$? Up to this point, I had only considered the pairs when $b \sim a$, and I thought that $``\text{it's time to move towards the cases where} \ b = 13a!"$ In sync with the Cauchy-Schwarz equality condition, when $3 \times 41 = 123 \sim 121 = 11^2$, I put a mental note that all terms must be proportionate with each other, and when I decided to drop $c$ a bit from 11 to 10, it would seem to be a good idea when I inductively descent-ed $(a,b,c)$ to $(a,b-1,c)$. However, this only worked for loose $ab-c^2 \geq a$. When it's just close enough, I must seek for another way in. \[\begin{tabular}{p{12cm}} $\rule{4cm}{0.5pt} \hspace{1cm} \bigstar \hspace{1cm} \rule{4cm}{0.5pt}$ \\ \end{tabular} \\ \]Back to proportionately close triples, the notice to take $(7,10,8)$ is clear enough: $7 \times 10$ is roughly equal to $8^2$. Then, I had the idea to let $c$ be a fixed variable and $a,b$ represented in terms of $c$ --- and of course, the triple $(c-1,c+2,c)$ got the pick. If this is the case, this can be done easily enough by setting $n = c-1$ and $y_1 = 2$ to propel $x_1y_1$ from being $1$ to being $2$. Interestingly, this way of thinking also brought me into the questioning of this configuration, as it's the first one to produce a quite general one which works for an infinite number of triples. That questioning brought forth an interesting way to approach the problem: instead of trying so hard to find $x_i$ and $y_i$, why don't I fix the $n$ or ($\#$ terms in each sequence) first --- and can I predict with clairvoyance what $n$ will be based on the values of $a,b,c$ alone? With this, in my judgement at that time, the only clear clue as to what to consider is the $``\text{discriminant}"$ $D = ab-c^2$. Surprisingly, this turned out to be a quite good predictor of $n$. Sure, when $D = 0$, $n$ could still be required to adapt with the squarefree part of $a$ and $b$, but otherwise, $n$ might as well be $1$. Moreover, for the relatively prime $(a,b,c)$ such as $(2,5,3)$ or $(8,11,9)$, $n$ is exactly $D+1$. Overall, $D$ seems to be quite inaccurate when $\gcd(x_1,x_2,\ldots,x_n) \ne 1$ such as in $(a,b,c) = (8,13,10)$ --- with that said, I still tried to ask the question: what if $a$ is actually a square, but its circumstances forced it to split into smaller parts? Considering $a = 4$, I tried to form a configuration which forced $\{x_i\} = \{1,1,1,1\}$; and I succeeded. \[\begin{tabular}{p{12cm}} $\rule{4cm}{0.5pt} \hspace{1cm} \bigstar \bigstar \hspace{1cm} \rule{4cm}{0.5pt}$ \\ \end{tabular} \\ \]From these advances alone, it would seem that no improvements are being made aside from considering the quantity $D$. After this, almost all attempts to induct $(a,b,c)$ was fruitless. The closest I've got is when $a$ is not the smallest number between $(a,b,c)$, implying $c$ to be the smallest one, I could let $(a,b,c) = (n,n+E+F,n+E)$, but any attempts of smoothing such as \[ n(n+E+F) \geq (n+E)^2 \Rightarrow nF \geq E(E+n) \]did not work as well as I thought it'd be. Since I only have those configurations I have crafted to my control, I thought I'd reviewed it back (as what I'm doing here) and noticed that $\textbf{not only does Fibonaccis exist on tight triples, but some form of it must exist in all!}$ Take $(3,41,11)$, the loosest pair $(a,b)$ I crafted. The illusion of proportionate numbers blinded me that the $(x_i,y_i)$ is equal to $(1,4),(1,4)$ and $(1,3)$, where $4 = 3+1$. When I dissected all pairs $(x_i,y_i)$ among all configurations, it would seem that most are in the form \[ (x_i,y_i) = (0,1), (1,1), (1,2), (2,3), (3,5), \ldots \]and the lowest $D$'s sequences (making the most unique sequences $\{x_i\},\{y_i\}$ among all configurations) are closely approximated by the golden ratio (also see that $(7,10,8)$ has two distinct sequences, the eighth and tenth). Only using the pairs $(1,1)$ and $(1,2)$, I further assembled them into the triples $(4,10,6)$ and $(4,13,9)$ --- speaking of which, this is the first time I made a triple based on its $\{x_i\},\{y_i\}$ isntead of solving for a suitable $\{x_i\}, \{y_i\}$. Analyzing the fourth last and third last triple, I noticed that $D$ is equal to $4$ and $3$, respectively. Now, I wondered what this could mean, and I realized that $D$ being $4$ might originate from $2 \times 2$, where there are two $(1,2)$ pairs and two $(1,1)$ pairs; and $3$ might similarly originate from $3 \times 1$. This prompted a long-forgotten manipulation of proving Cauchy-Schwarz as following: \begin{align*} D &= ab-c^2 = (x_1^2+x_2^2+x_3^2)(y_1^2+y_2^2+y_3^2) - (x_1y_1+x_2y_2+x_3y_3)^2 \\ &= (x_1y_2-x_2y_1)^2+(x_1y_3-x_3y_1)^2+(x_2y_3-x_3y_2)^2\\ \end{align*}and it all clicked. (And I hope that this made clear the long journey towards the apparently magical substitution in the Solution.) For further exposition, I continued making the second to last triple by combining $(2,3)$ and $(1,2)$ again, and say I would like to make $D$ constant when I inductively descended from $(9,22,14)$ to something else. To that end, I would like $x_iy_j-x_jy_i$ to stay constant throughout the process, and the easiest way to fix $x_iy_j-x_jy_i$, say, to be always equal to $1$ is to change all $y_i$ to be $y_i - k\cdot x_i$, as it would ensure that \[ x_i'y_j'-x_j'y_i' = x_i(y_j-k\cdot x_j)-x_j(y_i-k\cdot x_i) = x_iy_j-x_jy_i \]However, if we set $k = 2$ or greater, $c$ might decrease to $c-2a$, which we didn't quite want (but still doable, since $|c| < |a|$, but I digress) --- and more dangerously, we certainly do not want $b$ to change to $b+4a-2c$, which may be boundable but a bit uglier. I stuck with $k =1$ and computed $a',b',c'$, and there I was at the Solution. \[\begin{tabular}{p{12cm}} $\rule{4cm}{0.5pt} \hspace{1cm} \blacksquare \hspace{1cm} \rule{4cm}{0.5pt}$ \\ \end{tabular} \\ \]Final Comments: In theory, this problem really shouldn't be hard, and there are no big theoritical/soft tech leaps in this Solution and Motivation, but doing an old problem which combines a long-forgotten manipulation and tech might ramp up the difficulty a little. Overall, I would give this roughly 25M --- the first Section certainly contain a lot of unique configurations (and the desperation on how to induct this downwards was certainly very real), the second Section's hard tech being laughably obvious while located at a blind spot, and the final manipulation being a hard tech which may have been more well known in the past (USAMO 1999, anyone?).