If $(3p_n\pm \sqrt{5p_n^2-4})\Delta/(2p_n)$ is an integer then $p_n$ is an odd-indexed Fibonacci number

Solution 1:

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

2. 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 $L_{2m+1}$.

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

4. 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.