Rational number if n is even proof

**Jason Bourne** · January 13th 2009, 09:07 AM

Prove that the number

( $\sqrt{5} + 1)^n + (\sqrt{5} - 1)^n , (n \in \mathbb{N})$

is rational if and only if n is even.

(I think going one way if n is even then its clear that its rational, what about assuming that n is even then the number is rational? Is it inductive?)

**running-gag** · January 13th 2009, 09:37 AM

Hi

You can use Newton's formula to develop $(\sqrt{5} + 1)^n$ and $(\sqrt{5} - 1)^n$ then study the 2 cases (n even and n odd).

When n is even all the terms involving $\sqrt{5}$ disappear.

**Jason Bourne** · January 13th 2009, 10:17 AM

Newton's formula?

**running-gag** · January 13th 2009, 10:42 AM

Sorry I don't know how it is called in English
I am talking about this formula
$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k}\: a^k \:b^{n-k}$

**ThePerfectHacker** · January 15th 2009, 11:19 AM

Originally Posted by Jason Bourne

Prove that the number

( $\sqrt{5} + 1)^n + (\sqrt{5} - 1)^n , (n \in \mathbb{N})$

Consider the sequence $a_0,a_1,a_2,a_3,...$ defined as: $\left\{ \begin{array}{c} a_0 = a_1 = 2 \\ a_{n+2} = 2a_{n+1} + 24a_n \text{ for }n\geq 0 \end{array} \right.$

Certaintly, the terms of the sequence $\{ a_n \}$ are integers.
Furthermore, the solution to this recurrence relation is given by: $a_n = \left( 1 + \sqrt{5} \right)^n + \left( 1 - \sqrt{5} \right)^n$ .
If $n$ is even then $a_n = \left( 1 + \sqrt{5} \right)^n + \left( \sqrt{5} - 1 \right)^n \in \mathbb{N}$

Consider the Fibonacci sequence $f_0,f_1,f_2,...$ it satifies $f_n = \frac{1}{\sqrt{5}} \left[ \left( \frac{1+\sqrt{5}}{2} \right)^n - \left( \frac{1-\sqrt{5}}{2} \right)^n \right]$

Define $b_n = \left( 1+\sqrt{5}\right)^n + \left( \sqrt{5}-1\right)^n$ .

If $n$ is odd then $\frac{b_n}{2^n\sqrt{5}} = \frac{1}{\sqrt{5}} \left[ \left( \frac{1+\sqrt{5}}{2} \right)^n - \left( \frac{1-\sqrt{5}}{2} \right)^n \right] = f_n$ .

Therefore for odd $n$ we have $b_n = \sqrt{5}\cdot 2^n \cdot f_n\not \in \mathbb{Q}$ .

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