Paint the cells such that as many colors as possible are used. (i.e. for any two cells if they do not need to have the same color, then we paint them differently.) In such situation, I claim that we have used exactly $2^n$ colors. I will use induction. Consider the cells with both even coordinates. The configuration of their colors are in fact the same as the $2^{n-1}$ case. Thus $2^{n-1}$ colors are used for these cells by induction. And for a cell $(i,j)$ where at least one $i,j$ is odd, using the recurrence stated in the problem, we will return to $(i,j)$ after exactly $3\cdot 2^{n-1}$ cells. Thus there are $3\cdot2^{n-1}$ cells having the same color with $(i,j)$, including itself. This is in fact the last theorem in the paper http://www.ams.org/journals/mcom/1967-21-100/S0025-5718-1967-0223064-6/S0025-5718-1967-0223064-6.pdf Hence exactly $\frac{(2^n)^2\cdot\frac{3}{4}}{3\cdot2^{n-1}}=2^{n-1}$ colors are used for the remaining cells. The result follows.

Let $N = 2^n$. Consider the directed graph $G$ on $({\mathbb Z}/N)^2$ with edges $(j,i) \to (i+j, j)$. The problem reduces to showing that there are exactly $2^n$ connected components in this graph. First looking mod $2$, observe the vertices with both even coordinates are completely disjoint from those with both odd coordinates. This means that we'll proceed by induction on $n$, with base cases $n \le 2$ being done by hand. So we'll show that exactly $2^{n-1}$ new coordinates are generated on the $\frac34 \cdot N^2$ vertices with at least one odd coordinate (and applying induction hypothesis). The main structural claim is that: Claim: Let $a$ and $b$ be integers, at least one of which is odd. Then $(a,b)$ is contained in a cycle of length exactly $3 \cdot 2^{n-1}$. In the motivating example $(1,0)$ this implies the cycle is $$(1,0) \to (2,1) \to (3,2) \to (5,3) \to (8,5) \to \dots.$$ More generally, starting from $(a,b)$ the $(k+1)$st vertex is $(aF_{k+1} + bF_{k}, aF_{k} + bF_{k-1})$, where $F_0 = 0$, $F_1 = 1$, $F_2 = 1$, $F_3 = 2$, is the Fibanocci sequence. We will now prove three lemmas, from which the claim follows immediately by setting $(a,b) \equiv (aF_{k+1} + bF_k, aF_k + bF_{k-1})$ and doing calculation. (Again $(1,0)$ is the main motivating example.) Lemma: We have $F_k \equiv 0 \pmod 4$ if and only if $k \equiv 0 \pmod 6$. In that case, $F_{k-1} \equiv 1 \pmod 4$. Proof. Obvious, just write out the sequence modulo $4$. $\blacksquare$ Lemma: For all $m \ge 1$, $\nu_2(F_{6m}) = \nu_2(m) + 3$. Proof. The easiest proof is notice that $$F_{3m} = \frac{1}{\sqrt5} \left( \left( 2+\sqrt5 \right)^{2m} - (2-\sqrt5)^{2m} \right)$$ and then copy verbatim the proof of the exponent lifting lemma. $\blacksquare$ Lemma: For all $m \ge 1$, $\nu_2(F_{6m-1} - 1) = \nu_2(F_{6m+1} - 1) = \nu_2(m) + 2$. Proof. Let $a = F_{6m-1}$, $b = F_{6m}$, so $\nu_2(b) = \nu_2(m) + 3 \ge 3$. We use the identity $$F_{6m}^2 + 1 = F_{6m-1}F_{6m+1} \implies b^2 + 1 = a(a+b) \implies a = -\frac{1}{2} b + \underbrace{\sqrt{5(b/2)^2 + 1}}_{c}.$$ Now $c \equiv 1 \pmod 4$ since $a \equiv 1 \pmod 4$, and $$(c-1)(c+1) = \frac{5b^2}{4}.$$ Hence $\nu_2(c-1) = 2\nu_2(b)-2-1 \ge \nu_2(b)$. So $\nu_2(a) = \nu_2(b)-1$ as desired. $\blacksquare$ Finally this implies the problem since each cycle has length $3 \cdot 2^{n-1}$ and the number of squares is $\frac 34 N^2$, which gives $2^{n-1}$ cycles.

Define a color-sequence to be a sequence $a_0,a_1,...$ taken $\pmod{2^n}$ with $a_{n+2}=a_{n+1}+a_n$. The key is that the squares at $(a_n, a_{n+1})$ are the same for all $n$ given a color sequence $\{a_n\}$. Note that $\{a_n\}$ is periodic $\pmod{2^n}$, so we just need to prove that the number of equivalence classes of color-sequences $\pmod{2^n}$ is at most $2^n$. As an example, we have the color-sequences $1,0,1,1,2,3,\ldots$; $3,0,3,3,2,1,\ldots$; $0,0,\ldots$; and $2,2,0,\ldots$ $\pmod{4}$. First we prove a lemma: Lemma: Let an odd color-sequence be a color sequence with at most one of $a_0, a_1$ even. Then the period of an odd color-sequence is $3\cdot 2^{n-1}$. Proof: We inductively show that $\phi^{3\cdot 2^{n-1}}=(1+a\cdot 2^{n+1})+(b\cdot 2^n)\sqrt{5}$ for integers $a,b$ with $b$ odd. It is easy to check that for $n=2$ we have $\phi^6=9+4\sqrt{5}$. Now, if the result is true for $n$, note that $$\phi^{3\cdot 2^n}=[(1+a\cdot 2^{n+1})^2+5(b\cdot 2^n)^2] + [2(1+a\cdot 2^{n+1})\cdot (b\cdot 2^n)]\sqrt{5}$$ which is of the form $(1+a'\cdot 2^{n+2})+(b'\cdot 2^{n+1})\sqrt{5}$, as desired. Now use the extended $2$-adic valuation. We show that the least integer $d_n$ such that $v_2(\phi^{d_n}-1)=n$ is $d_n=3\cdot 2^{n-1}$. It is easy to check that $d_2=6$. Now, clearly $d_n\mid d_{n+1}$. But by the information above, $v_2\left(\phi^{3\cdot 2^{n-1}}-1\right)=n$, so it follows that $d_n\neq d_{n+1}$. Thus $d_{n+1}\ge 2d_n$, and it is easy to see that $v_2(\phi^{2d_n}-1)=n+1$, as desired. Now suppose $\{a_n\}$ was an odd color-sequence but has period less than $3\cdot 2^n$. Write $a_n=x\phi^n+y(-\phi)^{-n}$ and let $p$ be the period. We can easily check that the period is even by showing that the period is even for any such sequence $\pmod{4}$. Now, if $a_p\equiv a_0$ and $a_{p+1}\equiv a_1$, then $2^n\mid (\phi^p-1)(x-y\phi^{-p})$ and $2^n\mid (\phi^p-1)(x\phi+y\phi^{-p-1})$. But $v_2(\phi^p-1)<n$, so $2\mid x-y\phi^{-p}, x\phi+y\phi^{-p-1}$. But then it follows that $2\mid \phi(x-y\phi^{-p})-(x\phi+y\phi^{-p-1})=-y(\phi^{1-p}+\phi^{-1-p})$. Since $v_2(\phi)=0$, we get $2\mid y(\phi^2+1)=\sqrt{5}y\phi$. Thus $2\mid y$. Similarly, we get $2\mid x$. But then it follows $2\mid a_0,a_1$, and $\{a_n\}$ is not an odd color-sequence, contradiction! Thus the order of all odd color-sequences is $3\cdot 2^{n-1}$. $\blacksquare$ Now we prove the statement that there are $2^n$ equivalence classes of color-sequences $\pmod{2^n}$ through induction. Note that the statement is true for $n=2$ (as shown above). A non odd color-sequence $\pmod{2^n}$ is equivalent to a color-sequence $\pmod{2^{n-1}}$, since any sequence of form $2b_0,2b_1,...\pmod{2^{n}}$ has a corresponding sequence $b_0,b_1,...\pmod{2^{n-1}}$. There are $2^{n-1}$ such sequences in this case. Else, consider the odd color-sequences. There are $3\cdot 2^{2n-2}$ ways to choose the first two terms: $3$ ways to choose parity and $2^{n-1}$ for each parity. Now since any two consecutive terms are unique to each equivalence class, it follows that there are $\frac{3\cdot 2^{2n-2}}{3\cdot 2^{n-1}}=2^{n-1}$ equivalence classes, since there are $3\cdot 2^{n-1}$ unique pairs of consecutive terms per equivalence class by our lemma above. It follows that there are $2^{n-1}+2^{n-1}=2^n$ equivalence classes $\pmod{2^n}$, completing the inductive step. $\square$

Please check my dolution

Let $F_1=F_2=1,F_{n+2}=F_n+F_{n+1}$ be the Fibonacci sequence. It's easy to verify that $6\mid n \iff 4\mid F_n$ and that $F_{2n}= F_n(F_{n-1}+F_{n+1}), F_{2n+1}=F_n^2 + F_{n+1}^2$. (Both of these follow from the well-known $F_{m+n} = F_mF_{n+1} + F_{m-1}F_n$.) Lemma 1: $2^{k+1} \mid F_{3\cdot 2^{k-1}}$ for $k\ge 2$ and $F_{3\cdot 2^{k-1} +1} \equiv 2^k+1 \pmod {2^{k+1}}$ for all $k\ge 1$. Proof: We induct on $k$ with the appropriate base cases. The first statement's inductive step is trivial since $F_{3\cdot 2^k} = F_{3\cdot 2^{k-1}} (F_{3\cdot 2^{k-1}-1} + F_{3\cdot 2^{k-1}+1})$; the second term is the sum of two odd terms, so it contributes a factor of $2$. Similarly, the second statement follows because $F_{3\cdot 2^{k-1}+1} = F_{3\cdot 2^{k-2}}^2 + F_{3\cdot 2^{k-2}+1}^2$; the first term vanishes modulo $2^{k+1}$ for $k\ge 3$ while, by the inductive hypothesis, the second term is of the form $(a\cdot 2^k + 2^{k-1} + 1)^2 \equiv 2^k+1\pmod {2^{k+1}}$. Now I claim the period of $F_n$ modulo $2^k$ is $3\cdot 2^{k-1}$ for any $k\ge 1$. We can verify $k=1,2$ manually. For $k\ge 3$, let $p$ be the period. By the lemma, we know $F_{3\cdot 2^{k-1}},F_{3\cdot 2^{k-1}+1}\equiv 0,1\pmod {2^k}$, hence we can easily show $F_{n+3\cdot 2^{k-1}}\equiv F_n\pmod {2^k}$ so $p\mid 3\cdot 2^{k-1}$; meanwhile, since $F_{3\cdot 2^{k-2}+1} \equiv 2^{k-1} + 1\not\equiv 1\pmod {2^k}$ by the lemma, we see that $p\nmid 3\cdot 2^{k-1}$. Since $4\mid F_p$, our first observation yields $6\mid p$; in particular, since $3\mid p$, the above divisibilities tell us we must have $p=3\cdot 2^{k-1}$ as desired. Now we can easily prove the original problem by induction on $n$ with base cases $n=0,1$ easy to verify. For the inductive step, consider any value of $n>1$. Note that the operation $f(i,j) =(j,i+j)$ (taken modulo $2^n$) is clearly a bijection on the cells of the table. Now for any square $(a,b)$ with $2\mid a,b$, we see that $f^k (a,b)$ always has two even coordinates, so by dividing the coordinates by $2$, we get a correspondence to the squares of a $2^{n-1}\times 2^{n-1}$ table following the same constraints. Therefore, the $2^{2n-2}$ cells $(a,b)$ with two even coordinates can be colored by a maximum of $2^{n-1}$ colors; now we ignore such squares and focus on squares with at least one odd coordinate. In this case, note that $f^k (a,b) = (F_{k-1}a + F_kb, F_ka+F_{k+1}b)$. Because $f$ is a bijection, in any maximal coloring the squares with the same color correspond to a cycle of $f$; I claim this cycle always has length $3\cdot 2^{k-1}$. Indeed, a cycle requires $f^k(a,b)=(a,b)$, or $F_{k-1}a + F_kb \equiv a, F_ka + F_{k+1}b \equiv b$; since every third square in this cycle has two odd coordinates, WLOG suppose $a,b$ are both odd. Then it's not hard to show that we need $F_k\equiv 0, F_{k-1}\equiv F_{k+1}\equiv 1\pmod {2^n}$, meaning the size of the cycle is just the period of $F$ modulo $2^n$, or $3\cdot 2^{n-1}$. Now since there are $3\cdot 2^{2n-2}$ total squares with $(a,b)$ not both even, we get $2^{n-1}$ total cycles; combining with the previous observation gives $2^n$ total colors as desired.

From arguements similar to the above solutions, we just need that the Fibonacci sequence has period $3*2^{n-1}$ modulo $2^n$. Let $A= \begin{bmatrix} 0&1\\ 1&1 \end{bmatrix}$ Claim: $$v_2(\det(A^{3*2^{n-1}}-I))=2n.$$ Proof: Note that $$\det(A^{3*2^{n-1}}-I)=\det(A^{3*2^{n-2}}-I)\det(A^{3*2^{n-2}}+I)$$ by induction, $\det(A^{3*2^{n-2}}-I)$ has 2n-2 powers of 2 Now, we just need $\det(A^{3*2^{n-2}}+I)$ has only 2 powers of 4. This follows from the fact that $F_{3*2^{n-1}}+F_{3*2^{n-1}+2}=2mod4$.(Examine the fibonacci sequence mod 4, has period 6) Thus, our claim has been proven. Thus, $A^{3*2^{n-1}}-I$ cannot be the 0 map modulo $2^{n+1}$. By another arguement that has been repeated in above solutions, we can show by induction $A^{3*2^{n}}-I$ is the 0 map modulo $2^{n+1}$. Hence we are done.

Solution with william122 The cases $n = 0,1$ are direct to verify. Suppose now $n \ge 2$. Draw directed edge $(i,j) \to (j,i+j)$ for each square $(i,j)$. Define an odd square to be a square with at least one odd coordinate. Note that each cycle in the $n-1$ case through vertices $v_1,...,v_t$ directly bijects to a cycle in the $n$ case through vertices $2v_1,...,2v_t$. Observe that if a cycle contains an odd square then all squares in the cycle must be odd. We will show that for a $2^n\times 2^n$ board, 1) there are $2^{n-1}$ cycles through the odd vertices 2) for each cycle $C$ going through some odd square, the set of squares in the $2^{n+1} \times 2^{n+1}$ board that are in $C$ $\pmod{2^n}$ is covered by 2 cycles $C_1, C_2$ $\pmod 2^{n+1}$, with $|C_1| = |C_2| = 2|C| = 3\cdot 2^n$. For each cycle through squares $(s_1,s_2), (s_2,s_3),...,(s_t,s_1)$ define the corresponding sequence by $(s_0,s_1,s_2,...)$ with $s_i := s_{i-1} + s_{i-2} \forall i$. Note that a cycle passes through some odd square if and only if there are no consecutive even terms in the sequence $(s_i)$. There are $a,b \in \mathbb Z[\sqrt{5}]$ such that $s_i = a\alpha^i + b(\frac{-1}{\alpha})^i$ since $\alpha := \frac{1+\sqrt{5}}{2}, \frac{-1}{\alpha}$ are roots of the characteristic polynomial $x^2 - x- 1$. Lemma: The smallest $m$ for which there exists $\kappa \in \mathbb Z_2[\sqrt{5}]$ such that $\alpha^m = 1 + 2^n\kappa$ is $m = 3\cdot 2^{n-1}$ for all $n \ge 1$. Proof: Note that when $n = 1$, $m = 3$. By LTE we have $\nu_2(\alpha^{3m} -1) = 1 + \nu_2(m)$, which concludes the proof. $\blacksquare$ Now suppose $s_t, s_{t+1}\equiv s_0, s_1\pmod {2^n}$ respectively, then $$a\alpha^t + b(\frac{-1}{\alpha})^t \equiv a + b.$$ $$a\alpha^{t+1} + b(\frac{-1}{\alpha})^{t+1} \equiv a\alpha + b(\frac{-1}{\alpha})$$ We then obtain $a(\frac{-1}{\alpha} - \alpha)(\alpha^t - 1) = 0$, but $\frac{-1}{\alpha} - \alpha = -\sqrt{5}$, thus $a(\alpha^t - 1) = 0$. Likewise $b(\alpha^t - 1) = 0$. Thus if $\nu_2(\alpha^t -1) < n$ then $2|a,b$, contradiction to the sequence passing through an odd square. Thus any cycle passing through an odd square of an $2^n \times 2^n$ has length equal to the order of $\alpha \pmod {2^n}$ which is $3\cdot 2^{n-1}$. It suffices to show that each cycle splits into $2$, and we do this by noting that for each sequence $s$, $(s_{3\cdot 2^{n-2}},s_{3\cdot 2^{n-2}})$ is congruent to $(s_0,s_1) \pmod {2^{n-1}}$ yet not congruent $\pmod {2^n}$, hence their difference is a multiple of $2^{n-1}$. Note that for each odd square $(a,b) \pmod {2^{n-1}}$ there are $4$ squares $\pmod {2^n}$ (namely $\{a,a+2^{n-1}\}\otimes \{b, b+2^{n-1}\}$) congruent to it $\pmod {2^{n-1}}$, and for each sequence $s\pmod {2^n}$ congruent to some sequence $s' \pmod {2^{n-1}}$ that correspond to cycles on the odd squares, $s$ covers only $2$ of the $4$ residues produced by each square of the cycle corresponding to $s'\pmod {2^{n-1}}$. Since the cycles covering $s' \pmod {2^n}$ must be disjoint it follows that there are exactly $2$ of them. $$\mathcal{YUH}.$$

Let $(i,j)$ deontes the $j^{th}$ cell of the $i^{th}$ row. For any $a,b$, let the sequence with $$a_0=a,a_1=b,a_{n+2}=a_{n+1}+a_n$$ be the $(a,b)-$Fibonaaci sequence, where the terms are considered modulo $2^n-1$. From the condition, the cell formed by every two consecutive terms in the $(a,b)-$ Fibonacci sequence has the same color of $(a,b)$. Define the equivalence relation $(a_1,b_1)\sim(a_2,b_2)$ if $(a_1,b_1)$ belongs to the $(a_2,b_2)-$ Fibonacci sequence. It suffices to show that there are exactly $2^n$ equivalence classes. We proceed by induction on $n$. The base cases are trivial. Now suppose all smaller case holds, we consider the case $n$. Now suppose $(a,b)$ is a cell in the case $n-1$. Then $(2a,2b)$ is a cell in the case $n$. Meanwhile, the equivalence classes of $(a,b)$ in the case $n-1$. has the same number of elements as the equivalence classes of $(a,b)$ in the case $n$. Hence there're $2^{n-1}$ equivalence classes which contains $(a,b)$ with both $a$ and $b$ even. Now we show the following: CLAIM. If $a$ and $b$ are not both even, then the equivalence classes containing $(a,b)$ has $3\cdot 2^{n-1}$ elements. Or in other words, the $(a,b)$-Fibonacci sequence has period $3\cdot 2^{n-1}$. Proof. Let $\{a_n\}$ be the $(a,b)-$Fibonacci sequence Notice that let $F_n$ be the $n^{th}$ Fibonacci number then $$a_{n+2}=F_{n+1}a_1+F_na_0$$ Hence it suffices to show that the Pisano period of Fibonacci sequence mod $2^n$ is $3\cdot 2^{n-1}$. This is well-known but I will still prove it. We claim that for $n\geq 2$, $$v_2(F_{3\cdot 2^{n-1}})=n+1$$ and $$F_{3\cdot 2^{n-1}-1}\equiv 2^{n}+1 \pmod{2^{n+1}}$$ We proceed by induction on $n$. The base cases are trivial. Suppose these are true for some $n-1$, we consider the case $n$. Indeed, we have $$F_{3\cdot 2^{n}}=F_{3\cdot 2^{n-1}}(F_{3\cdot 2^{n-1}-1}+F_{3\cdot 2^{n-1}+1})$$ By the inductive hypothesis we have $F_{3\cdot 2^{n-1}-1}+F_{3\cdot 2^{n-1}+1}\equiv 2\pmod{2^n}$ Hence $v_2(F_{3\cdot 2^{n}})=v_2(F_{3\cdot 2^{n-1}})+1=n+1$. Now by inductive hypothesis we have $$F_{3\cdot 2^{n-1}-2}\equiv 2^n-1\pmod{2^{n+1}}$$ Hence $F_{3\cdot 2^{n-1}-2}=2^{n+1}k+2^n-1$ for some odd $k$. Therefore $$F_{3\cdot 2^n-2}=(2^{n+1}k+2^n-1)^2=2^{n+1}-1\pmod{2^{n+2}}$$ From which we easily obtain $$F_{3\cdot 2^{n}-1}=F_{3\cdot 2^{n}}-F_{3\cdot 2^{n}-2}=2^{n+1}+1\pmod{2^{n+2}}$$ $\blacksquare$ Since there are $2^2n-2^{2n-2}=3\cdot 2^{2n-2}$ cells whose coordinates are not all even, hence they form $2^{n-1}$ equivalence classes, this completes the proof.

An equivalent formulation is to consider the ring $R_n=\mathbb{Z}[\varphi]/2^n\mathbb{Z}[\varphi]$ (which is isomorphic to $\mathbb{Z}/2^n\mathbb{Z}[\varphi]$), where $\phi=\frac{1+\sqrt{5}}{2}$, and to consider the number of orbits obtained by repeatedly multiplying by $\varphi$. This is because $(a+b\varphi)\varphi=a\varphi+b\varphi^2=a\varphi+b+b\varphi=b+(a+b)\varphi$, which corresponds to passing from the cell $(a,b)$ to $(b,a+b)$. Since $\varphi$ is invertible in this ring, it suffices to count the number of such orbits in $R_n^\times$, and add this number from $1$ to $n$, and also add the orbit due to $0\cdot \varphi^n$. Therefore, if we find the order of $\varphi$ $ord_n(\varphi)$, the number of orbits will be $\frac{|R_n^\times|}{ord_n(\varphi)}$. Lemma: $ord_n(\varphi)=3\cdot 2^{n-1}$ Proof: for $n=1$, it is trivial, and in fact we have $v_2(\varphi^3-1)=v_2(1+2\varphi-1)=v_2(2\varphi)=1$ (we define $v_2(x)$ for $0\neq x\in \mathbb{Z}[\varphi]$ as the highest power of $2$ dividing both coefficients). So we can now use LTE (the proof is as normal and basically comes down to prove that $v_2(x^{2^k}+y^{2^k})=1$ if $x\equiv y\pmod{2}$; clearly $x^{2^k}+y^{2^k}\equiv 2x^{2^k}\equiv 0\pmod{2}$, and if it was also true mod $4$ we would have $\left(\frac{x^{2^{k-1}}}{y^{2^{k-1}}}\right)^2 \equiv -1$, but this is not possible, because $(a+b\varphi)^2=(a^2+b^2)+(2ab+b^2)\varphi \equiv -1\pmod{4}\implies a^2+b^2\equiv -1\pmod{4}$ which is a contradiction), for $n\geq 1$ because $2|\varphi^3-1$ and $2^n$ is even. We therefore obtain: $v_2(\varphi^{3\cdot 2^{n}}-1)=v_2(\varphi^6-1)-1+v_2(3\cdot 2^n)=v_2(5+8\varphi-1)+n-1=2-1+n=n+1$, which implies $ord_n(\varphi)=3\cdot 2^{n-1}$ for $n\geq 1$[by induction we have $3\cdot 2^{n-1}=ord_n(\varphi)|ord_{n+1}(\varphi)|3\cdot 2^n$, but $ord_{n+1}(\varphi)$ can't be $3\cdot 2^{n-1}$ since $v_2(\varphi^{3\cdot 2^{n-1}}-1)$ is only $n$ and not $n+1$]. Finally the number of orbits in $R_n^\times$ must be $\frac{|R_n^\times|}{ord_n(\varphi)}=\frac{(2^n)^2-(2^{n-1})^2}{3\cdot 2^{n-1}}=\frac{3\cdot 2^{2(n-1)}}{3\cdot 2^{n-1}}=2^{n-1}$. Therefore, summing this number for all $k=1,...,n$ and including the contribution of the orbit of $0$, we get that the total number of orbits, and thus the maximal number of colours is $$1+\sum_{k=1}^n 2^{k-1}=1+2^n-1=2^n$$ as wanted.

$+\!\!\!\!\!\!\:\times$ Sketch: We can reduce the problem to finding the period of the Fibonacci sequence $\!\!\!\mod{2^n}$ by considering the cycles $$(i,j) \to (j,i+j) \to (i+j,i+2j) \to \cdots.$$ Let $F_n$ be the $n$th Fibonacci number. We prove two claims: Claim: If $6 \mid n$, then $\nu_2(F_{2n})=\nu_2(F_n)+1$. Proof: Let $L_n$ be the $n$th Lucas number. Use $F_{2n}=F_nL_n$ and take $L_n \!\!\mod{8}$. Claim: $\nu_2(F_{3 \cdot 2^{n+1}})=n+1$ for all nonnegative integers $n$. Proof: Use the fact that $F_{2n+1}=F_n^2+F_{n+1}^2$. Combining these claims gives the desired result.

Wrong.

Correspond the cells to $2$-dimensional vectors in $({\mathbb Z}/2^n{\mathbb Z})^2$. Let $A= \begin{bmatrix} 0&1\\ 1&1 \end{bmatrix}$. We want to show that the digraph $\textbf{v} \to A\textbf{v}$ has exactly $2^n$ connected components. (Thus, one may color the unfriendly cells in the equality case suggested by inductive hypothesis, and color each cycle with a single unique color, for a total of $2^n$ colors.) We proceed by induction on $n \ge 2$ (since $n=1$ is workable by hand), with the base case trivial. For the inductive step, call cells with both coordinates even unfriendly and all other cells friendly, and note that they are all part of connected components independent of that of all other cells. Furthermore, scaling down by a factor of $2$ implies that there are exactly $2^{n-1}$ unfriendly connected components, by inductive hypothesis on the $n-1$ case. The key claim is the following. Claim: The graph of friendly vertices is compromised of cycles of length exactly $3 \cdot 2^{n-1}$. Proof. We show that $A^{3 \cdot 2^{n-1}} \equiv I$ modulo $2^n$. More specifically, we have $$\nu_2(\det(A^{3 \cdot 2^{n-1}} - I)) = n$$ which we will prove; this will show that $3 \cdot 2^{n-1}$ is indeed the smallest period that works. We induct on $n \ge 2$ with base case easy. Note that $$\det(A^{3 \cdot 2^{n-1}} - I) = \det(A^{3 \cdot 2^{n-2}} - I)\det(A^{3 \cdot 2^{n-2}} + I).$$ By inductive hypothesis, $2^{n-1} \parallel \det(A^{3 \cdot 2^{n-2}} - I)$, so it suffices to show that $2 \parallel \det(A^{3 \cdot 2^{n-2}} + I)$, or rather that this determinant is $2 \pmod{4}$. Note that (inductively), $A^k$ has a main diagonal of elements that are both $F_k$, the $k$th Fibonacci number. It is easy to see that the $3 \cdot 2^{n-2}$nd Fibonacci number is even (by say, mod $3$ bash), so $\det(A^{3 \cdot 2^{n-2}} + I)$ is just $$\det(A^{3 \cdot 2^{n-2}}) + 2 \cdot F_{3 \cdot 2^{n-2}} + 1 \equiv 1 + 0 + 1 \equiv 2 \pmod{4},$$ and we are done. Thus, there are at exactly $$\frac{(2^n)^2-(2^{n-1})^2}{3 \cdot 2^{n-1}} = 2^{n-1}$$ friendly connected components. Adding the unfriendly connected components, there are exactly $2^n$ connected components in total.

I have obtained a miraculous proof of this fact which doesn't use the pisot period of $F_n$. So it's probably a fakesolve but idk how plz check. DM me if you find the fake part of it. Note that this gives chains. $$(i, j) \to (i+j, i) \to (2i+j, i+j) \to \dots$$ It remains to show that there are at most $2^n$ such chains. Note that these chains end up being pairs of elements from the Fibonacci sequence. Claim: There are exactly $2^n$ such chains. Proof. Consider the residues $\pmod{2}$ of the first element. If the element is $(0, 0)$, then it follows that by mapping $(i, j) \to \left(\frac{i}{2}, \frac{j}{2}\right)$ that there are $2^n$ of these pairs. If the element is $(0, 1), (1, 0), (1, 1)$, shift till its $(1, 1)$. Then subtract the chain generated by $(1, 1)$ from the original chain, and divide each element of tha chain by $2$. This gives us our other $2^n$ pairs. $\blacksquare$