Show that $y^n+1/y^n \in \mathbb{N}$

Question

Let $y$ be the solution of the equation $x+\frac{1}{x}=3$ such that $y>1$.

Show that $y^n+\frac{1}{y^n} \in \mathbb{N}$

We have $x+\frac{1}{x}=3$ if and only if $ x^2-3x+1=0$.

Which has two solutions:

$x_1=\frac{-b-\sqrt{b^2-4ac}}{2a}= \frac{3-\sqrt{5}}{2}$

$x_2=\frac{-b+\sqrt{b^2-4ac}}{2a}= \frac{-3+\sqrt{5}}{2}$

I don't observe an $y > 1$ what is the problem?

Edit :

So after I fixed my error ! I need to show that :

$(\frac{-3+\sqrt{5}}{2})^n +( \frac{2}{-3+\sqrt{5}})^n$ is a natural number ! Can someone give me hint !

Answer 1

This problem is best solved with recurrences. Let $f(n) = y^n + \frac{1}{y^n}$, and you know $f(1)=1$. Find f(2) with some manipulation, then find a recursion that relates $f(n)$ to $f(n-1)$ and $f(n-2)$. You should be able to see that f(n) is an integer for all n.

Answer 2

Your error is that $b= -3$ so $-b = -(-3)= 3$.