Fibonacci Arithmetic

 (Dylan here.)

For this post (and perhaps my subsequent blog posts), I will share a math-adjacent slice of my experience, and then talk some (probably unrelated) math. I will also include upfront some problems related to the non-math/math I am talking about, in case you want to try them first.

• There are 650 points inside a circle of radius 16. Prove there exists a ring with inner radius 2 and outer radius 3 covering 10 of these points.
• Prove that a graph on $2n$ vertices with more than $n^2$ edges contains $n$ triangles.
• Let $F_0=0,F_1=1,F_{n+2}=F_{n+1}+F_n$ be the Fibonacci numbers. Prove that $F_{n}$ divides $F_{2n}$.
• When does $n$ divide $F_n$?

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A. Math-Adjacent

I got into maths olympiad pretty early (upper primary), and via puzzles. Not the Sudoku/Kakuro/Rush Hour type of puzzles (whose freshness quickly waned), but more like those from books such as The Canterbury Puzzles (Dudeney) or Perplexing Puzzles and Tantalizing Teasers (Gardner). Here's some of them (unfortunately dissected from any intriguing backstories, which I have long forgotten):

• A drawer has 10 black and 10 white socks. How many socks to I need to draw in the dark to guarantee a matching pair?
• (Battle of Hastings) Solve $a^2=13b^2+1$ in the positive integers. (Hopefully this reminds you of the previous blog post by Jit!)
• A grass pasture can feed $4$ cows for $4$ days, or $3$ cows for $6$ days. How many days can it feed $2$ cows for?
• Dissect an $8\times 8$ square into $4$ congruent pieces such that each piece contains exactly $1$ of the $4$ upper-left diagonal squares, and exactly $1$ of the $4$ upper-right diagonal squares.
• Find a second solution to $a^3+b^3=9c^3$ in the positive integers.

At first, I treated each problem as its own mini-puzzle, expecting a unique idea that leads to a solution. But unlike puzzle books where solutions were just a flip-to-the-back away, I spent a considerable amount of time struggling to solve olympiad problems. This was beneficial in a couple of ways:

• Every new problem was a test of whether I had internalised the ideas from past problems to be able to apply them in a fresh context.
• Learning a solution isn't from scratch, but to fill in the gap between what I've tried and what works.
• It had the psychological benefit of reinforcing delayed gratification, allowing me to spend more and more time on problems without losing focus or hope. (My relatives are often surprised at how I can sit $4.5$-hour contests. They don't know that I typically spend up to a week with a problem at the back of my mind.)
• Once in a while, coming up with a fresh idea that cracks a problem gives the most amazing sense of intellectual accomplishment, irreplicable in other aspects of (school) life. 

One of the early realisations that I had was that there was "different levels of understanding". For instance, I could confidently state the Pigeonhole principle soon after learning it, yet be clueless on how to apply it to subsequent problems. On the other hand, people who memorised $12\times 13=156$ were surprised on how I could compute $72\times 73 = 4900 + 350 + 6 = 5256$ in my head, perhaps reflecting that I had a more natural understanding of the distributivity law. Yet I was strangely unable to understand, let alone apply, the Inclusion-Exclusion principle, perhaps because the corresponding distributivity law in set theory was unintuitive to me: $A\cap (B\cup C) = (A\cap B) \cup (A\cap C)$.

In particular, I realised that understanding solutions line by line was completely different from understanding the essence and/or motivation. When peers or seniors present their solutions, correctness is typically prioritised, and questions like "But why did you consider this quantity/extremal choice?" were usually left unaddressed, or if prompted, passed over with replies such as "A similar problem used this idea before" or "It was just one of the many things I tried that finally worked". There was a certain subculture that the more problems you've seen/done, the better you are at maths. Which was true at a certain level of understanding, but not at the level I was interested in attaining. 

(Admittedly, that was not the only reason I didn't start grinding problems. My long term memory was pretty poor, and I often found myself not recognising problems I've seen, but still being able to re-solve them with time. Also, I was somehow comfortable with sitting in the struggle, not having a sense of inferiority for not knowing the solution. I had enough aha moments along the way to reward me for not giving up early. And perhaps most pivotally, there was little to no external pressure from parents/teachers/peers to grind.)

So, I kept my attention to fewer problems, but really thought hard about them, continually accumulating "What's the motivation?" questions somewhere in the recesses of my mind, with some faith that I would eventually learn/see the intuition. This turned out to not only be true, but also give way to even deeper levels of understanding (such as "How did someone set this problem in the first place?" or "Why is the theory set up/used in this way?"). 

Okay, that's enough of a slice for now. Onwards to maths! 

B. Fibonacci Arithmetic

So, I want to prove that $F_n$ divides $F_{2n}$. 

First things first (and you'll realise this question is more and more important as you grow older, perhaps in correlation with laziness): Why do we care? 

I defer its answer (and the answer to "How did someone come up with this problem?") to the end of the post. For now, let's ask the next best question: "Why do we even believe this to be true?"

B(i). Experiment, conjecture, induct

Therein lies the motivation for trying small cases. Let's compute the small Fibonacci numbers (or pull them from the same place in your brain that has $\pi$ to the first however many number of digits): 

$$n \quad 0 \quad 1 \quad 2 \quad 3 \quad 4 \quad 5 \quad 6 \quad 7 \quad 8 \quad 9 \quad 10$$

$$F_n \quad 0 \quad 1 \quad 1 \quad 2 \quad 3 \quad 5 \quad 8 \quad 13 \quad 21 \quad 34 \quad 55$$

$$F_{2n} \quad 0 \quad 1 \quad 3 \quad 8 \quad 21 \quad 55 \quad 144 \quad 377 \quad \dots \quad $$

It certainly is true that $F_n$ divides $F_{2n}$ for $n = 0,1,\dots,7$. In fact, let's compute the quotient:

$$F_{2n}/F_n \quad \text{nil} \quad 1 \quad 3 \quad 4 \quad 7 \quad 11 \quad 18 \quad 29 \quad \dots \quad $$

Notice anything? The quotient seems to obey its own "Fibonacci relation", where each successive term is the sum of the previous two! It seems natural to conjecture that $F_{2n} = F_n L_n$ where $L_0=2, L_1=1,L_{n+2}=L_{n+1}+L_n$. (Tangential remarks: the sequence $L_n$ in fact has its own name: the Lucas numbers. And, bonus points if you came to ask why $L_0=2$ is the "correct" resolution to the undefined quotient $0/0$, and/or what does it mean!)

At this point, the olympiad-trained minds might be screaming: "just try to induct already!" However, the more mathematically mature might also quickly realise that something is missing: trying to prove $F_{2n}=F_n L_n$ via a $2$-step induction $P(n-2)\land P(n-1)\implies P(n)$ is (given current progress) impossible, since (for instance) it doesn't differentiate this from the false statement $F_{2n} =^! F_n F_n$ (that is coincidentally true for $n=0,1$).

What do we do? I see two options:

• Perhaps run a larger induction, such as $P(n-3)\land P(n-2)\land P(n-1)\implies P(n)$, or strong induction.
• Since $L_n$ satisfies the same recurrence relation as $F_n$, perhaps the two sequences may be related in some way that makes the induction work.

In either case, we also need to establish a recurrence on $F_{2n}$ to run the induction. Let's do that first, and see where that leads us:

$$F_{2n} = F_{2n-2} + F_{2n-1} = F_{2n-2} + (F_{2n-2} + F_{2n-3}) = 2F_{2n-2} + (F_{2n-2}-F_{2n-4}) = 3F_{2n-2} - F_{2n-4}.$$

One may use this once more to express $F_{2n}$ in terms of $F_{2n-2},F_{2n-4},F_{2n-6}$, but the coefficients won't be unique. So, unless we know exactly what we're looking for, a larger induction is not natural at the moment.

Let's try to express what we need to show:

$$F_nL_n =^? 3F_{n-1}L_{n-1} - F_{n-2}L_{n-2},$$

and then try to simplify what we have to show with some manipulations:

$$F_nL_n =^? 3F_{n-1}L_{n-1} - F_{n-2}(L_n-L_{n-1}).$$

$$(F_n+F_{n-2})L_n =^? (3F_{n-1}+F_{n-2})L_{n-1}.$$

$$(F_n+F_{n-2})L_n =^? (2F_{n-1}+F_{n})L_{n-1}.$$

$$(F_n+F_{n-2})L_n =^? (F_{n-1}+F_{n+1})L_{n-1}.$$

At this point, it takes a sharp eye to realise that this implies

$$\frac{L_n}{F_{n-1}+F_{n+1}} =^? \frac{L_{n-1}}{F_n+F_{n-2}},$$

which means this fraction is in fact constant in $n$. Inputting small values of $n$, it crystallises that in fact $L_n = F_{n-1}+F_{n+1}$; this is precisely the relationship between $L_n$ and $F_n$ that we needed! And since $L_n$ and $F_n$ satisfy the same recurrence relation, so this statement itself is true by a simple induction on $n$. 

Let's now write the proof in the forward direction: 

1. First, establish the relation $F_{2n} = 3F_{2n-2} - F_{2n-4}$.
2. Next, induct on statement $P(n): F_{2n} = F_n(F_{n-1}+F_{n+1})$.
3. Check base cases $P(1),P(2)$ and prove $P(n-2)\land P(n-1)\implies P(n)$ as above.

(Hindsight remark: the identity $L_n=F_{n-1}+F_{n+1}$ was reverse-engineered from the conjectured identity $F_{2n} = F_nL_n$. However, this could have been obtained systematically: for instance, noting that $(L_0,L_1) = (2,1) = 2(1,1) - (0,1) = 2(F_1,F_2)-(F_0,F_1)$, this should extend by induction to $L_n = 2F_{n+1} - F_n = F_{n+1}+F_{n-1}$, as desired.)

(Meta-remark: this is an example of when the presented solution is not instructive to hinting or motivating a solution. While the "key step" in this presented solution appears to be staking the claim $F_{2n} = F_n(F_{n-1}+F_{n+1})$, it entirely obscures the entire exploration between the first step $F_{2n} = F_nL_n$ to understanding the quotient $L_n$ well enough to express it as $F_{n+1}+F_{n-1}$. In particular, a student who "arrived at $L_n = F_{n+1}+F_{n-1}$ by clever trial and error" would perhaps have been positively encouraged that "clever trial and error" is the way to go.) 

Now, this may have satisfied a young me, thinking that the takeaway from this problem is an extra identity of the Fibonacci numbers ($F_{2n} = F_nL_n, L_n = F_{n+1}+F_{n-1}$), and the two conjectures along the way were "motivated by small cases". But a slightly older me would have a ton of questions left unanswered:

• Did someone come up with this by computing the Fibonacci numbers and observing patterns?
• What's so special about the Fibonacci numbers that such identities exist, or such divisor relations are true? For instance, why doesn't $L_n$ divide $L_{2n}$ in general?
• Isn't it rather amazing that a "linear-type" recurrence relation $F_{n+2} = F_{n+1}+F_n$ relates to a "quadratic-type" identity $F_{2n}  = F_n(F_{n+1}+F_{n-1})$?
• Are there other identities? Does $F_{3n}$ relate to $F_n$?

Because our first solution wasn't that enlightening, let's do the only thing we can do: solve it again.

B(ii). Linear recurrence and Binet (algebra view)

A consequence of mathematics pop culture is that most of us know that the Fibonacci sequence is related to the golden ratio $\varphi = \frac{1+\sqrt 5}{2}$. Perhaps accompanied by pictures of an aesthetic spiral, we might know that the ratio $F_{n+1}/F_n$ tends to $\varphi \approx 1.618$ as $n$ is large. The underlying reason is that the golden ratio is involved in an explicit formula for the Fibonacci numbers (which I recently learnt to have a name: the Binet formula): 

$$F_n = \frac{\varphi^n - (-\varphi)^{-n}}{\sqrt 5} = \frac{1}{\sqrt 5}\left[\left(\frac{1-\sqrt 5}{2}\right)^n-\left(\frac{1-\sqrt 5}{2}\right)^n\right]$$

Now, when I first learnt of this, I was quite surprised:

• How can there be an exact formula involving $\sqrt 5$ for an integer-valued quantity? 
• How did anyone come up with this in the first place?

I soon saw how the $\sqrt 5$ component always vanished (since alternate terms of the binomial expansions of $(1\pm \sqrt 5)^n$ cancel), but the fact that the $2^n$ denominator always works itself out is completely unobvious to me.

If you are the type of person that cannot move on without a proof, see this website (https://artofproblemsolving.com/wiki/index.php/Binet%27s_Formula). Importantly, Binet's formula is a special case of a general result concerning closed form solutions of linear recurrences. This next paragraph will only make sense to those who already know the theory of linear recurrences:

The recurrence $F_{n+2} = F_{n+1}+F_n$ has characteristic equation $s^2 = s+1$, which has roots $\alpha,\beta= \frac{1\pm \sqrt 5}{2}$. The general solution is given by $F_n = A\alpha^n + B\beta^n$, and initial conditions $F_0=0,F_1=1$ determine the coefficients $A,B$, which give Binet's formula.

But the Lucas numbers $L_n$ also satisfy the same recurrence, so we expect a similar closed form! (Meta-remark: this type of realisation is not too accessible to those who memorise Binet's formula as one of several identities to be used as a tool, without truly understanding its context in the general theory of linear recurrences. Unfortunately, this is also how I feel when I use Pascal's theorem in Euclidean geometry.) Applying the initial conditions $L_0=2,L_1=1$ to solve for the coefficients gives 

$$L_n = \varphi^n + (-\varphi)^{-n}.$$

But this is exactly the desired quotient:

$$\frac{F_{2n}}{F_n} = \frac{\varphi^{2n} - (-\varphi)^{-2n}}{\varphi^{n} - (-\varphi)^{-n}} = \varphi^n + (-\varphi)^{-n} = L_n,$$

following from the all-too-familiar algebraic identity $x^2-y^2 = (x-y)(x+y)$! (This alludes to the reason $L_0 = 2$: because $\alpha^0 + \beta^0 = 1 + 1 = 2$. It also explains why non-linear relations may arise from a linear recurrence: because the closed form expression of $F_n$ involves exponentials in $n$.)

Okay, so we now have a second proof. Things are starting to get clearer. In fact, we can attempt to apply our tools to relate $F_{3n}$ to $F_n$:

$$\frac{F_{3n}}{F_n} = \frac{\varphi^{3n} - (-\varphi)^{-3n}}{\varphi^{n} - (-\varphi)^{-n}} = \varphi^{2n} + \varphi^n (-\varphi)^{-n} + (-\varphi)^{-2n} = L_{2n} + (-1)^n$$

so we also have that $F_n$ divides $F_{3n}$! 

(Those who are already comfortable with such manipulations should try to show that $F_n$ divides $F_{mn}$, and find an expression for the quotient $F_{mn}/F_n$.)

Yet, this proof has also opened up its own can of worms:

• What are irrational numbers like $\sqrt 5$ doing in a proof of a purely number-theoretic statement about a purely integer-valued integer-recurrence sequence?

Well, time to solve it again.

B(iii). Local analysis (number theory view)

This time, we'll stick to the integers. The nice thing about recurrences is that they are still recurrences modulo $m$ for any $m$. For example, modulo $3$, the Fibonacci sequence goes

$$0\quad 1 \quad 1 \quad 2 \quad 0 \quad 2 \quad 2 \quad 1 \quad 0 \quad 1 \quad 1 \quad 2 \quad 0 \quad 2 \quad 2 \quad 1 \quad \dots$$

looping with period $8$. But the fact that it loops is no surprise: there are only finitely many values modulo $3$, and any consecutive two values uniquely determine the sequence not only after (via $F_{n+2}=F_{n+1}+F_n$), but also before (via $F_n = F_{n+2}-F_{n+1}$). It follows from a Pigeonhole argument that modulo $m$, the Fibonacci sequence loops with period $\leq m^2$.

Okay, but we are interested in showing that $F_{2n}$ is a multiple of $F_n$, so let's take modulo $F_n$. The Fibonacci sequence goes 

$$0\quad 1 \quad 1 \quad 2 \quad 3 \quad \dots\quad F_{n-2} \quad F_{n-1} \quad 0 \quad F_{n-1} \quad F_{n-1} \quad 2F_{n-1} \quad 3F_{n-1} \quad \dots $$

Wait, but that means we are done! An induction that $F_{n+k} \equiv F_k F_{n-1} \pmod{F_n}$ for $k=0,1,\dots,n$ suffices. In fact, this proves the more general statement that $F_{n}$ divides $F_{mn}$! 

(Technical remark: this is not truly a "local"-type proof. To work locally in number theory is to work modulo primes $p$, then modulo prime powers $p^k$ Hensel's-Lifting-Lemma style, then finally modulo $m$ Chinese-Remainder-Theorem-style.)

Again, questions:

• Can I say anything about the period of the Fibonacci sequence modulo $m$? For instance, can I characterise exactly when $n$ divides $F_n$?
• Does the final induction statement hint at an identity $F_{n+k} = F_kF_{n-1} + G_kF_n$ for some $G_k$?  

Let's go again. 

B(iv). Counting (combinatorics view)

Don't we all love enumerative combinatorics: the Catalan numbers and such. Learning that the Fibonacci number $F_n$ counts the number of ways to tile a $1\times (n-1)$ rectangle with $1\times 1$ and $1\times 2$ pieces can only suggest that there is a combinatorial interpretation of the identity $F_{2n} = F_n(F_{n+1}+F_{n-1})$, and thus a combinatorial proof that $F_n$ divides $F_{2n}$.

How do we relate the number of ways to tile a $1\times (2n-1)$ rectangle to the number of ways to tile a $1\times (n-1)$ rectangle? The specific dimensions pretty much encourage you to divide the $1\times (2n-1)$ rectangle in half, i.e. split cases by the tile covering the centre square. In each case, the rectangle is divided into two smaller rectangles, one of which has dimensions $1\times (n-1)$. Since the number of valid tilings of a pair of disjoint rectangles multiply, thus $F_n$ divides the number of ways to tile each case, thus $F_n$ divides $F_{2n}$ as desired!

Explicitly unpacking the discussion above, the tile covering cell $n$ is either of the three possible cases $\{n\}, \{n-1,n\}, \{n,n+1\}$. The total number of tilings is thus

$$F_{2n} = F_{n}F_{n} + F_{n-1}F_n+F_nF_{n-1} = F_n(F_n + 2F_{n-1}) = F_nL_n$$

which is what we previously concluded! 

Let's try to generalise our argument to show $F_n$ divides $F_{mn}$. To count the number of ways to tile a $1\times (mn-1)$ rectangle, let's run a case check on the tiles covering cells at positions $n,2n,\dots,(m-1)n$. This divides the rectangle into $m$ smaller rectangles, and crucially in all cases, at least one piece is of dimensions $1\times (n-1)$. Thus $F_n$ divides $F_{mn}$ as desired!

Okay, now let's try to get an explicit quotient. We may do so by considering the three cases of the tile covering cell $n$. But actually, we can be slightly clever and group two of the cases together, so it is now "is the boundary between cells $n-1$ and $n$ covered by a domino, or not?". This gives the relation

$$F_{n+k} = F_{n-1}F_k + F_nF_{k+1}$$

which is something we conjectured from the previous section! This also gives us a way of expressing the quotient $F_{mn}/F_n$ inductively: 

$$\frac{F_{mn}}{F_n} = F_{(m-1)n+1} + F_{n-1}\frac{F_{(m-1)n}}{F_n}.$$

Okay, the critical question:

• This feels like reverse engineering to cook up a proof. Does this proof interact in any way with the number-theoretic/algebraic perspectives?

Okay, enough alternative solutions for now. Let's tie up some loose ends.

B(v). Reconciling algebra with number theory

It's time to answer the question of "how do I reconcile real numbers like $\varphi$ with the discrete world of number theory?"

The answer is simpler than it seems: just allow exactly the numbers that you need! Unlike weird numbers like $\pi$ or $e$, the golden ratio satisfies the nice polynomial equation $\varphi^2=\varphi+1$. This means it is algebraic, which means that including it in your "world of allowable numbers" is less scary than you might think! Let's illustrate this with an example:

Beginning with the integers (in which the Fibonacci sequence naturally lives), suppose you decide that you also allow arithmetic (plus, minus, times) using the number $\varphi$. But that means we also allow things like $2+\varphi$, or $3-5\varphi$. But we also have $(2+\varphi)(3-5\varphi)$, or $(2+\varphi)^n$. 

But let's expand and simplify using the "nice polynomial equation satisfied by $\varphi$":

$$(2+\varphi)(3-5\varphi) = 6 -7\varphi -5\varphi^2 = 6-7\varphi -5(\varphi+1) = 1-12\varphi.$$

Hey, that's not too bad! It seems like by adding $\varphi$ to our world of allowable numbers, our initial terrifying picture of the space

$$\{a_0 + a_1\varphi + \dots + a_n\varphi^n : a_0,a_1,\dots, a_n \in \mathbb Z\}$$

turns out to simplify to "$2$ integer coordinates", or "integer-dimension $2$":

$$\{a+b\varphi : a,b\in \mathbb Z\}$$

with the "multiplication rule"

$$(a_1+b_1\varphi)(a_2+b_2\varphi) = (a_1a_2+b_1b_2) + (a_1b_2+a_2b_1+b_1b_2)\varphi.$$

Where does the Fibonacci sequence naturally lie in this space? The Fibonacci relation $F_{n+2} = F_{n+1}+F_n$ is precisely captured by "multiplying by $\varphi$": 

$$(a+b\varphi)\varphi = b + (a+b)\varphi,$$

or more explicitly,

$$(F_{n}+F_{n+1}\varphi)\varphi = F_{n+1} + F_{n+2}\varphi.$$

From this and abit of exploration, one realises that "the Fibonacci numbers are the coefficients of powers of $\varphi$ when written in the standard basis $\{1,\varphi\}$":

$$\varphi^n = F_{n-1} + F_n\varphi.$$

But in this world, one may then easily relate $F_n$ to $F_{2n}$:

$$\varphi^{2n} = F_{2n-1} + F_{2n}\varphi = (\varphi^n)^2 = (F_{n-1} + F_n\varphi)^2 = (\text{integer}) + (2F_{n-1}F_n+F_n^2)\varphi,$$

implying the identity we have proven over and over again: $F_{2n} = F_n(2F_{n-1}+F_n) = F_nL_n$.

(Meta-remark: if you're like me, and learnt about rings like the Gaussian integers $\mathbb Z[i]$, or perhaps $\mathbb Z[\sqrt 2]$ in the context of Pell's equations, then your first instinct (especially after seeing the Binet formula) might not be "to add $\varphi$", but to "add $\sqrt 5$", then "add powers of $2$ in the denominator", then begin to worry about the non-UFD nature of $\mathbb Z[\sqrt5]$. You might be pleasantly surprised that "adding $\varphi$" circumvents this mess, and all it took was to free oneself from the shackles of the precise explicit elements of Binet's formula, or realise that $\sqrt 5$ is not any "nicer" as a number than $\varphi$, or some other indescribable internal realisation.)

(Second meta-remark: picturing $\mathbb Z[\varphi]$ as a $2$-dimensional lattice is perfectly fine! You might be thinking: "$\varphi$ is a real-number, so any drawing of $\mathbb Z[\varphi]$ must live in the $1$-dimensional real line, and can't be pictured $2$-dimensionally". After thinking long and hard about this, my (imperfect) response is: this precisely reflects the subtle yet jarring difference between "$\mathbb Z$-dimension" and "$\mathbb R$-dimension". It is as if the concepts live in different worlds. The apparent conflict then comes from the fact that we depict a lattice by aligning $\Z$-dimensions with geometric dimensions, which preserves some notions such as addition, but distorts other notions such as size and total ordering inherited from the reals. For instance, $-727+690\varphi$ and $260+80\varphi$ are very far apart in the lattice picture, although their real values agree to $6$ significant digits. The fact that they are "far apart in number theory" suggests the lattice picture is appropriate for the current problem.)

Trailing remarks for this subsection: if we only care about the Fibonacci numbers modulo $3$, one may work in the world of numbers $\{a+b\varphi : a,b\in \mathbb F_3\}$ where $\mathbb F_3$ is the integers-modulo-$3$ field. One may also see that it is equivalent to working in $\{a+b\sqrt 5 : a,b\in \mathbb F_3\}$, or $\{a+bi : a,b\in \mathbb F_3\}$.

However, if we work modulo $11$, note that $s^2=s+1$ has roots $s\equiv 4,8$ modulo $11$. So, we may identify $\varphi \equiv  4$ (or $\sqrt 5 \equiv -4$, say). We then have a translated Binet formula:

$$F_n \equiv 8(4^n-8^n) \pmod {11}.$$

B(vi). Linear algebra (opening up linear recurrence theory)

This post is getting long, but I wanted to mention a final perspective. Unfortunately, it involves matrices, and the default blog environment isn't too friendly with that. So I'll use the semicolon $;$ to denote the next line in the matrix, so $(a; b)$ is a $2\times 1$ column vector.

The key fact is that the matrix $A = (1 \quad1; 1\quad 0)$ encapsulates the Fibonacci recurrence. This is because

$$\pmatrix{F_{n+2} \\ F_{n+1}} = \pmatrix{1 & 1 \\ 1 & 0} \pmatrix{F_{n+1} \\ F_n}$$

Now, this might satisfy some of the older readers, and turn off the younger readers who have only learnt the rules of matrix addition and multiplication without any context to its application. Nevertheless, once stated in this form, and with some additional knowledge in certain areas of mathematics, one may draw certain links:

• (Linear algebraists) The golden ratio and its counterpart arise naturally as the eigenvalues of the matrix $A$. Binet's formula is what you get upon diagonalising $A$. 
• (Number theorists) We can skip being careful with $(\Z/m\Z)[\varphi]$, and just work with $A$ and the elements modulo $m$; for instance, $$A^n = \pmatrix{F_{n+1} & F_n \\ F_n & F_{n-1}} \equiv \pmatrix{\ast & 0 \\ 0 & \ast}\pmod {F_n}$$
• (Physicists) If linear recurrences are the discrete version of a differential equation, then the Fibonacci recurrence is a dynamical system analogous to a second-order differential equation (naively $y'' = y' +y$, but not really). The matrix relation above is then analogous to interpreting the second-order differential equation as a first-order equation in more variables, by working in phase space (naively $(y,y')$). 

I feel like the physicists' view is the most natural motivation for the matrix identity, but I'm sure plenty of others will justifiably disagree. 

Trailing tangential remark for this subsection: the Fibonacci numbers also arise as the coefficients of a power series:

$$\frac{x}{1-x-x^2} = 0 + 1x + 1x^2 + 2x^3 + 3x^4 + 5x^5 + \dots + F_n x^n+\dots$$

Ardent fans of analysis might try to try to find an alternate proof to $F_{2n} = F_nL_n$ by manipulating this series/using a contour integral to extract the $x^n$-coefficient/other advanced methods.

B(vii). Have we finally understood the problem?

To close the discussion, let's return to the two important meta questions: How did someone come up with this problem? and, Why do we care?

Think back to the first solution B(i) we obtained, by trying small cases, making empirical conjectures, and proving them by induction. Now think at the rest of the different approaches and ideas that came after. Hopefully it is now clear that while solution B(i) may have been the most direct way to solve the problem, the other approaches were probably the more natural way to think up/discover/create the problem. While I cannot say for sure which particular method best inspires investigating the relationship between $F_n$ and $F_{2n}$, here's a Fibonacci problem of my own: 

$$\text{(ICMC '23 R1) Prove that }3^{2023}\text{ divides }3^2F_4 +  3^3F_6  +  3^4F_8  +  \dots  +  3^{2023}F_{4046}.$$

The final form of the problem as presented above, or its most natural solution(s), has absolutely nothing to do with the initial ideas that inspired it: identities involving the logarithm and arctangent power series.

Now for the most touchy question: Why do we care? Perhaps the answer lies in the fact that the problem itself ("$F_n$ divides $F_{2n}$") is only one part of the journey; looking above, we have gone much further than the problem itself to

• Develop/use general tools in linear recurrence theory, ring extensions, and linear algebra.
• Go on to conjecture and prove other divisor relations (e.g. $F_n$ divides $F_{mn}$) or identities (e.g. $F_{n+k}  = F_nF_{k+1} + F_{n-1}F_k$).
• Ask new questions that go beyond extending the proof techniques for the current problem (e.g. when does $n$ divide $F_n$?).

Perhaps this reflects that we understood-understood this problem, or at the very least, learnt a bunch more cool maths around it. And hopefully there's some aspect(s) of the cool maths above that interests you, and/or helps you pose and tackle harder problems to come.

C. Conclusion

In the personal narrative, I talked about puzzles as my entrypoint into maths olympiad, the benefits of struggling with a problem (instead of, say, reading solutions), the realisation that there are different levels of understanding for maths, and why I didn't prioritise quantity of problems while doing maths. In the maths section, I talked about hopefully enough different ways to attack a simple-at-first-glance problem, adding layer upon layer of insight to rival Kuih Lapis. The meta-point I wanted to bring home was that maths doesn't end once the problem is solved, but I guess I also got carried away and wrote a blog post three times longer than I initially expected. If you're still here, thanks for reading!

Trailing remarks, for real this time:

• What can you say about $\gcd(F_m,F_n)$?
• "$\mathbb Z[i]$ multiplication" $(a+bi)(c+di) = (ac-bd)+(ad+bc)i$ precisely encodes the trigonometric addition formulae. Does multiplication in $\mathbb Z[\varphi]$ encode analogous addition formulae (perhaps involving some other pair of functions/objects)?
• (For the old people) There are nice values taken by infinite sums such as $\sum \frac{1}{1+F_{2n+1}}$ or $\sum \frac{1}{F_{2^n}}$. But why would we expect them to be nice in the first place/where do they naturally arise?