# On $4n+1 = x^2, 5n+1 = y^2$ and the Fibonacci numbers

While answering this question, I decided to look at the particular case,

$$4n+1 = x^2\\5n+1 = y^2$$

to be solved simultaneously. The solutions for $n,x,y$ are A157459, A007805, and A049629, respectively,

\begin{aligned} n_k &= 72, 23256, 7488432, 2411251920,\dots\\[2.5mm] x_k &= \frac{F(6k+3)}{2}\\ &= 17, 305, 5473, 98209, 1762289, 31622993,\dots\\[2.5mm] y_k &= \frac{-F(6k+1)+F(6k+5)}{4} = \frac{F(6k+2)+F(6k+4)}{4}\\ &= 19, 341, 6119, 109801, 1970299, 35355581,\dots \end{aligned}

starting with $k=1$ and $F(k)$ are the Fibonacci numbers.

Trying to find an expression for the $n_k$ in terms of the Fibonacci numbers, the only remaining possibilities are $F(6k)$ and $F(6k+6)$. After some fiddling, I found,

$$n_k = \frac{-18+\phi^{12k+6}+\phi^{-(12k+6)}}{80} = \frac{F(6k)\,F(6k+6)}{16}\tag1$$

and where $\phi = \frac{1+\sqrt{5}}{2}$ is the golden ratio.

Q: How do we prove that the two expressions for $(1)$ are indeed equivalent?

The Fibonacci numbers are given in terms of $\phi$ by $$F(n) = \frac{\phi^n - (-\phi)^{-n}}{\sqrt{5}}.$$ Also note that for even $n$, $(-\phi)^{-n} = \phi^{-n}$. This implies that \begin{align*} F(6k) F(6k+6) &= \frac{1}{5} (\phi^{6k} - \phi^{-6k})(\phi^{6k+6} - \phi^{-6k - 6}) \\ &= \frac{1}{5} \left[ \phi^{12k + 6} + \phi^{-(12k + 6)} - \phi^6 - \phi^{-6} \right] \\ &= \frac{1}{5} \left[ \phi^{12k + 6} + \phi^{-(12k + 6)} - 18 \right], \end{align*} since $\phi^6 + \phi^{-6} = 18$. The given equality then follows.
EDIT: Just to prove that $\phi^6 + \phi^{-6} = 18$: note that $$(\phi^6 + \phi^{-6})(\phi^6 - \phi^{-6}) = \phi^{12} - \phi^{-12},$$ from which it follows that $$\phi^6 + \phi^{-6} = \frac{F(12)}{F(6)} = \frac{144}{6} = 18.$$
• Thanks to your pointing out that $\phi^6+\phi^{-6} = 18$, I found a short formula for the general case $kn+1 = x^2,\, (k+1)n+1 = y^2$. (I had puzzled how to express the constant in the numerator.) Kindly see the updated answer in this post. – Tito Piezas III Jan 22 '16 at 2:21