2
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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 $$ \phi^6 + \phi^{-6} = \frac{F(12)}{F(6)} = \frac{144}{6} = 18. $$

$\endgroup$
  • $\begingroup$ Ahh, very nice. $\endgroup$ – Tito Piezas III Jan 22 '16 at 1:17
  • $\begingroup$ 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. $\endgroup$ – Tito Piezas III Jan 22 '16 at 2:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.