Solve without calculator!

Similar to Arpan Mathur 's approach.

Note that $\varphi = \dfrac {1+\sqrt 5}2$ is the golden ratio. Hence $\dfrac 1\varphi = \dfrac 2{1+\sqrt 5} = \dfrac {\sqrt 5-1}2$ $\implies \varphi - \dfrac 1\varphi = 1$ . Then we have:

$\begin{aligned} \left(\varphi - \frac 1\varphi \right)^2 & = 1^2 \\ \varphi^2 - 2 + \frac 1{\varphi^2} & = 1 \\ \varphi^2 + \frac 1{\varphi^2} & = 3 \\ \left(\varphi - \frac 1\varphi \right) \left(\varphi^2 + \frac 1{\varphi^2} \right) & = 1\times 3 \\ \varphi^3 - \varphi + \frac 1\varphi - \frac 1{\varphi^3} & = 3 \\ \varphi^3 - \frac 1{\varphi^3} & = 3+1 = 4 \\ \left(\varphi^3 - \frac 1{\varphi^3}\right)^2 & = 16 \\ \varphi^6 - 2 + \frac 1{\varphi^6} & = 16 \\ \varphi^6 + \frac 1{\varphi^6} & = 16 + 2 = 18 \\ \left(\varphi^3 - \frac 1{\varphi^3}\right) \left(\varphi^6 + \frac 1{\varphi^6}\right) & = 4 \times 18 \\ \varphi^9 - \varphi^3 + \frac 1{\varphi^3} - \frac 1{\varphi^9} & = 72 \\ \varphi^9 - \frac 1{\varphi^9} & = 72 + 4 = 76 \\ \left(\varphi^6 + \frac 1{\varphi^6}\right) \left(\varphi^9 - \frac 1{\varphi^9}\right) & = 18 \times 76 \\ \varphi^{15} + \varphi^3 - \frac 1{\varphi^3} - \frac 1{\varphi^{15}} & = 1368 \\ \varphi^{15} - \color{#3D99F6}\frac 1{\varphi^{15}} & = 1368 - 4 = 1364 & \small \color{#3D99F6} \text{Since } -1 < \frac 1{\varphi^{15}} < 0 \\ \left \lfloor \varphi^{15} \right \rfloor & = \boxed{1364} \end{aligned}$

Let $\alpha = \frac{1 + \sqrt{5}}{2}$ and $\beta = \frac{1 - \sqrt{5}}{2}$ ; We proceed to show that $\alpha^{15} + \beta^{15}$ is an integer.

Notice that $\alpha$ and $\beta$ are the roots of the quadratic equation $x^2 - x -1 = 0$ i.e

$\alpha^2 = \alpha\ +1 \quad...(1)$ and $\beta^2 = \beta+ 1 \quad ...(2)$ . Multiplying $(1)$ by $\alpha^{n-2}$ and $(2)$ by $\beta^{n-2}$ for some integer $n \geq 3$ we get:

$\alpha^n = \alpha^{n-1} +\alpha^{n-2}$ and $\beta^n = \beta^{n-1} +\beta^{n-2} \Rightarrow \alpha^n + \beta^n = (\alpha^{n-1} + \beta^{n-1}) + (\alpha^{n-2} + \beta^{n-2})$

Define $L_k = \alpha^{k} + \beta^{k}$ for natural $k$ . Now $L_1 = 1, L_2 = 3$ and $L_n = L_{n - 1} + L_{n -2}$ for $n\geq 3$ .(This sequence is called Lucas Sequence.)

The sequence so formed is $1,3,4,7,11,18,29,47,76,123,199,322,521,843,1364,2207,...$ whose 15th term is 1364. Hence $L_{15} = \alpha^{15} + \beta^{15} = 1364$ .

Now, $\alpha^{15} = 1364 - \beta^{15} \Rightarrow \lfloor \alpha^{15}\rfloor = \lfloor1364 + (- \beta^{15})\rfloor = 1364 + \lfloor-\beta^{15}\rfloor$

Also $0 < -\beta < 1 \Rightarrow 0 < -\beta^{15} < 1 \Rightarrow \lfloor-\beta^{15}\rfloor = 0 \Rightarrow \lfloor \alpha^{15}\rfloor = 1364$ .