If $(3p_n\pm \sqrt{5p_n^2-4})\Delta/(2p_n)$ is an integer then $p_n$ is an odd-indexed Fibonacci number Suppose $1<p_1<\dots<p_n$ are pairwise relatively prime integers and let $\Delta=p_1\cdots p_n$. In the proof of Lemma 4.8 of this paper (https://arxiv.org/pdf/1606.08656.pdf), there is the following paragraph.
If $$D=\frac{3p_n\pm \sqrt{5p_n^2-4}}{2}\cdot \frac{\Delta}{p_n} $$ is an integer then $p_n$ is an odd-indexed Fibonacci number. In this case $\sqrt{5p_n^2-4} $ is an odd-indexed Lucas number and $D=p_1\cdots p_{n-1}F_{2m-1}$. Note that $D=\frac{3p_n- \sqrt{5p_n^2-4}}{2p_n}\Delta<\frac{2}{3}\Delta$.
Actually the paper is about symplectic embeddings of rational homology balls into $\Bbb CP^2$, and I can't understand most of this paragraph. I only know the definitions of Fibonacci and Lucas numbers.

*

*Why is $p_n$ an odd-indexed Fibonacci number $F_{2m+1}$ (assuming $D$ is an integer)?


*In this case why is $\sqrt{5p_n^2-4} $ an odd-indexed Lucas number?


*Also how do we have $D=p_1\cdots p_{n-1}F_{2m-1}$?


*In the last sentence how the sign $\pm$ in $D$ became $-$?
 A: 

*

*Why is $p_n$ an odd-indexed Fibonacci number $F_{2m+1}$ (assuming $D$ is an integer)?





*In this case why is $\sqrt{5p_n^2-4} $ an odd-indexed Lucas number?


If $D$ is an integer, then $5p_n^2-4$ has to be a perfect square. It is known that if either $5N^2-4$ or $5N^2+4$ is a perfect square, then $N$ is a Fibonacci number. Since one can prove by induction that $5F_n^2+4(-1)^n=L_n^2$, one can say that $p_n$ is an odd-indexed Fibonacci number $F_{2m+1}$, and that $\sqrt{5p_n^2-4}$ is an odd-indexed Lucas number $L_{2m+1}$.



*Also how do we have $D=p_1\cdots p_{n-1}F_{2m-1}$?





*In the last sentence how the sign $\pm$ in $D$ became $-$?


If we choose $+$, then we have
$$D=\frac{3p_n+\sqrt{5p_n^2-4}}{2}\cdot \frac{\Delta}{p_n}=\bigg(\frac 32+\frac{\sqrt{5p_n^2-4}}{2p_n}\bigg)\Delta\gt \Delta$$
which contradicts $D\le\Delta$.
So, we have to choose $-$, and we finally get
$$\begin{align}D&=\frac{3p_n-\sqrt{5p_n^2-4}}{2}\cdot \frac{\Delta}{p_n}
\\\\&=\frac{3F_{2m+1}-\sqrt{5F_{2m+1}^2-4}}{2}\times (p_1\cdots p_{n-1})
\\\\&=\frac{3F_{2m+1}-L_{2m+1}}{2}\times (p_1\cdots p_{n-1})
\\\\&\color{red}=\frac{3F_{2m+1}-(2F_{2m+2}-F_{2m+1})}{2}\times (p_1\cdots p_{n-1})
\\\\&=(2F_{2m+1}-F_{2m+2})\times (p_1\cdots p_{n-1})
\\\\&=\bigg(2F_{2m+1}-(F_{2m+1}+F_{2m})\bigg)\times (p_1\cdots p_{n-1})
\\\\&=(F_{2m+1}-F_{2m})\times (p_1\cdots p_{n-1})
\\\\&=F_{2m-1}\times (p_1\cdots p_{n-1})\end{align}$$
where the fact that $L_n=2F_{n+1}-F_n$ (which can be proved by induction) was used in the equality in red.
