Solution to POTW #207: Evaluate x^{16}-987x for \sqrt{2}<x<\sqrt{3}

[anemone]

Here is this week's POTW:

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Let $a$ be a positive number and we show decimal part of the $a$ with $\{a\}$.

For a positive number $x$ with $\sqrt{2}<x<\sqrt{3}$ such that \(\displaystyle \{\frac{1}{x}\}=\{x^2\}\), evaluate $x^{16}-987x.$

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[anemone]

Congratulations to the following members for their correct solution::)

1. kaliprasad
2. greg1313
3. lfdahl

Solution from kaliprasad:

Spoiler

Because $ x> 1$ so $\frac{1}{x} < 1$ and hence $\lbrace \frac{1}{x} \rbrace = \frac{1}{x}$
Because $\sqrt{2} < x < \sqrt{3}$ hence $ 2 < x^2 < 3$ hence $\lbrace x^2 \rbrace = x^2 - 2$
hence $\frac{1}{x} = x^2-2$
or $ x^3-2x = 1$ or $(x+1)(x^2-x-1) = 0$ or $x^2 = x + 1$ as $x > 0$
$=>x^4 = (x+1)^2 = x^2 + 2x + 1 = (x+1)+(2x+1) = 3x + 2$
$=>x^8 = (3x+2)^2 = 9x^2 + 12 x + 4 = 9(x+1) + 12 x + 4 = 21x + 13$
$=>x^{16} = (21x+13)^2 = 441x^2 + 546 x + 169 = 441(x+1) + 546 x + 169 = 987 x + 610$
or $x^{16} - 987 x = 610$

Alternate solution from lfdahl:

Spoiler

Let $d$ denote the decimal part of $x^2$: $x^2 = 2+d = 2+\left \{ x^2 \right \} = 2+\left \{ \frac{1}{x} \right \}$.

$\left \{ x^2 \right \} = \left \{ \frac{1}{x} \right \} \Rightarrow d = \frac{1}{\sqrt{2+d}} \Rightarrow d^3+2d^2-1 = 0$

$d = -1$ is one of the roots in the cubic expression. Polynomial division yields:

\[d^3+2d^2-1 = (d+1)(d^2+d-1)\]

The set of solutions is:

\[d \in \left \{ -1, \frac{\sqrt{5}-1}{2}, \frac{-\sqrt{5}-1}{2} \right \}\]

From $d = \frac{1}{x} = \frac{\sqrt{5}-1}{2}$, we get: $x = \frac{\sqrt{5}+1}{2} $

So the $x$ in question is the famous golden ratio, $\varphi$.

One of the beautiful relations of the golden ratio involves Fibonacci numbers:

$\varphi^n = F_n\varphi + F_{n-1}$

In order to find $\varphi^{16}$, we simply use the 15th and 16th Fibonacci number:

\[\left \{ \underbrace{1}_{F_1},\underbrace{1}_{F_2},2,3,5,8,13,21,34,55,89,144,233,377,\underbrace{610}_{F_{15}}, \underbrace{987}_{F_{16}},1597, 2584 . . . \right \}\]

$\varphi^{16}=987\varphi + 610$

Thus we end with the result: $\varphi^{16}- 987\varphi = F_{15}= 610$.