Bay Area Mathematical Olympiad

Solution:
Let $\Phi=\frac{1+\sqrt{5}}{2}$ and $\varphi=\frac{1-\sqrt{5}}{2}$. Note for later use that $\Phi \varphi=-1, \Phi-\varphi=\sqrt{5}, \Phi=\Phi^{2}-1$, and $\varphi=\varphi^{2}-1$. We use Binet's formula for the Fibonacci numbers: $F_{n}=\frac{1}{\sqrt{5}}\left(\Phi^{n}-\varphi^{n}\right)$. (The reader who is not familiar with this formula may prove it inductively by checking that it works for $n=1,2$ and is compatible with the Fibonacci recurrence.)
Each $P_{n}$ may be written as $F_{j} F_{k}$ with $j \geq k$. Binet's formula gives
$$\begin{aligned} F_{j} F_{k} & =\frac{1}{5}\left(\Phi^{j}-\varphi^{j}\right)\left(\Phi^{k}-\varphi^{k}\right) \\ & =\frac{1}{5}\left(\Phi^{j+k}+\varphi^{j+k}-\Phi^{j} \varphi^{k}-\Phi^{k} \varphi^{j}\right) \\ & =\frac{1}{5}\left(\Phi^{j+k}+\varphi^{j+k}-(\Phi \varphi)^{k}\left(\Phi^{j-k}+\varphi^{j-k}\right)\right) \\ & =\frac{1}{5}\left(\Phi^{j+k}+\varphi^{j+k}-(-1)^{k}\left(\Phi^{j-k}+\varphi^{j-k}\right)\right) \\ & =\frac{1}{5}\left(L_{j+k}-(-1)^{k} L_{j-k}\right), \end{aligned}$$
where we define $L_{n}=\Phi^{n}+\varphi^{n}$. In what follows, we will use two properties of $L_{n}$ : it is positive for all $n \geq 0$, and $L_{n+4}>L_{n}$ for all $n \geq 0$. Both properties are easily proved via the observation that $L_{n}$ is, for all $n \geq 2$, the integer closest to $\Phi^{n}$.
Now fix $r$ and consider the set of products $F_{j} F_{k}(j \geq k)$ for which $j+k=r$. All of these products share a "leading" term of $\frac{1}{5} L_{r}$. The remaining term can be written as $-\frac{(-1)^{k}}{5} L_{r-2 k}$. By the two properties of $L_{n}$ noted above, we have
$$-L_{r-4}<-L_{r-8}<-L_{r-12}<\cdots<-L_{r-4\lfloor r / 4\rfloor}<L_{r-2-4\lfloor(r-2) / 4\rfloor}<\cdots<L_{r-10}<L_{r-6}<L_{r-2}$$
and thus
$$F_{r-2} F_{2}<F_{r-4} F_{4}<F_{r-6} F_{6}<\cdots<F_{r-5} F_{5}<F_{r-3} F_{3}<F_{r-1} F_{1} .$$
We note that the smallest and largest products in inequality (1) are $F_{r-2} F_{2}=F_{r-2}$ and $F_{r-1} F_{1}=F_{r-1}$, respectively. Thus the largest product $F_{j} F_{k}$ with $j+k=r$ is equal to the smallest product $F_{j} F_{k}$ with $j+k=r+1$. This implies that the sequence $P_{1}, P_{2}, P_{3}, \ldots$ consists of chains of the form (1) strung end to end for successively increasing values of $r$. All that remains is to show that the difference between any two consecutive terms in (1) is a Fibonacci number.
Such differences are of the form $\frac{1}{5}\left(L_{n+2}-L_{n-2}\right)$ (for some integer $n$ ), except in the middle where there is one difference of the form $\frac{1}{5}\left(L_{n+1}+L_{n-1}\right)$. We now show that both of these expressions are equal to $F_{n}$ :
$$\begin{aligned} F_{n} & =\frac{1}{\sqrt{5}}\left(\Phi^{n}-\varphi^{n}\right) \\ & =\frac{\Phi-\varphi}{5}\left(\Phi^{n}-\varphi^{n}\right) \\ & =\frac{1}{5}\left(\Phi^{n+1}+\varphi^{n+1}-\Phi \varphi\left(\Phi^{n-1}+\varphi^{n-1}\right)\right) \\ & =\frac{1}{5}\left(\Phi^{n+1}+\varphi^{n+1}+\Phi^{n-1}+\varphi^{n-1}\right) \quad\left(=\frac{1}{5}\left(L_{n+1}+L_{n-1}\right)\right) \\ & =\frac{1}{5}\left(\Phi^{n+2}-\Phi^{n}+\varphi^{n+2}-\varphi^{n}+\Phi^{n}-\Phi^{n-2}+\varphi^{n}-\varphi^{n-2}\right) \\ & =\frac{1}{5}\left(\Phi^{n+2}+\varphi^{n+2}-\Phi^{n-2}-\varphi^{n-2}\right) \quad\left(=\frac{1}{5}\left(L_{n+2}-L_{n-2}\right)\right) . \end{aligned}$$
Therefore, every term of $D_{1}, D_{2}, D_{3}, \ldots$ is a Fibonacci number.