Solution to a puzzle by @Numerus_A

The Puzzle

Compute the value of

$\sqrt{{(L_{2 n} + 2 \sqrt{\frac{6}{5}})}^{2} + F_{2 n}^{2}} - \sqrt{{(L_{2 n} - 2 \sqrt{\frac{6}{5}})}^{2} + F_{2 n}^{2}} (n \in N)$

Original Tweet

Solution

Let $a \geq 0$ and $b \geq 0$ .
$\begin{aligned} f_{a, b} (x) & := \sqrt{(a + x)^{2} + b^{2}} - \sqrt{(a - x)^{2} + b^{2}} \\ g_{a, b} (x) & := \frac{x}{2} \sqrt{\frac{a^{2} + b^{2} - x^{2} / 4}{a^{2} - x^{2} / 4}} \end{aligned}$

$\forall x \in [0, 2 a] : f_{a, b} (g_{a, b} (x)) = x$

Assume $x \in [0, 2 a]$ . Because $f_{a, b} (x) \geq 0$ , we may rewrite

$f_{a, b} (g_{a, b} (x)) = \sqrt{f_{a, b} {(g_{a, b} (x))}^{2}} = \sqrt{2} \sqrt{a^{2} + b^{2} + g_{a, b} {(x)}^{2} - \sqrt{(a^{2} + b^{2} + g_{a, b} {(x)}^{2})^{2} - 4 a^{2} g_{a, b} {(x)}^{2}}}$

$g_{a, b} {(x)}^{2} = \frac{(4 a^{2} + 4 b^{2} - x^{2}) x^{2}}{4 (2 a + x) (2 a - x)}$

$a^{2} + b^{2} + g_{a, b} {(x)}^{2} = \frac{16 a^{4} + 16 a^{2} b^{2} - x^{4}}{4 (2 a + x) (2 a - x)}$

$\sqrt{{(a^{2} + b^{2} + g_{a, b} {(x)}^{2})}^{2} - 4 a^{2} g_{a, b} {(x)}^{2}} = \frac{16 a^{4} + 16 a^{2} b^{2} - 8 a^{2} x^{2} + x^{4}}{4 (2 a + x) (2 a - x)}$

Plugging in these values into the above expression for $f_{a, b} (g_{a, b} (x))$ yields

$f_{a, b} (g_{a, b} (x)) = \sqrt{2} \sqrt{\frac{8 a^{2} x^{2} - 2 x^{4}}{4 (2 a + x) (2 a - x)}} = x$

$L_{2 n}^{2} = 5 F_{2 n}^{2} + 4$

We can use Binet-type expressions for Lucas and Fibonacci numbers:
$\begin{aligned} 5 F_{2 n}^{2} + 4 & = 5 {(\frac{ϕ^{2 n} - ϕ^{- 2 n}}{\sqrt{5}})}^{2} + 4 \\ = ϕ^{4 n} - 2 + ϕ^{- 4 n} + 4 \\ = ϕ^{4 n} + 2 + ϕ^{- 4 n} \\ = {(ϕ^{2 n} + ϕ^{- 2 n})}^{2} \\ = L_{2 n}^{2} \end{aligned}$
Alternatively we can also compute in the $Q$ -algebra $Q [\frac{1}{2}, \frac{\sqrt{5}}{2}]$ , as in a previous tweet by @Numerus_A :
$\begin{aligned} 1 & = ϕ^{- 2 n} ϕ^{2 n} \\ = \frac{L_{2 n} - F_{2 n} \sqrt{5}}{2} \frac{L_{2 n} + F_{2 n} \sqrt{5}}{2} \\ = \frac{L_{2 n}^{2} - 5 F_{2 n}^{2}}{4} \end{aligned}$

$g_{L_{2 n}, F_{2 n}} (4) = \frac{4}{2} \sqrt{\frac{L_{2 n}^{2} + F_{2 n}^{2} - 4^{2} / 4}{L_{2 n}^{2} - 4^{2} / 4}} = \frac{4}{2} \sqrt{\frac{6 F_{2 n}^{2}}{5 F_{2 n}^{2}}} = 2 \sqrt{\frac{6}{5}}$

そうして
$4 = f_{L_{2 n}, F_{2 n}} (g_{L_{2 n}, F_{2 n}} (4)) = f_{L_{2 n}, F_{2 n}} (2 \sqrt{\frac{6}{5}}) = \sqrt{{(L_{2 n} + 2 \sqrt{\frac{6}{5}})}^{2} + F_{2 n}^{2}} - \sqrt{{(L_{2 n} - 2 \sqrt{\frac{6}{5}})}^{2} + F_{2 n}^{2}}$

投稿日：2021年2月19日

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putputlaw

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putputlaw

Solution to a puzzle by @Numerus_A

The Puzzle
Solution