Problem: Show that whenever $n ≥ 3, f(n) > α^{n−2}$, where $α = (1 + \sqrt{5})/2$. $F(n)$ is the Fibonacci sequence $(1, 1, 2, \dots )$
Solution: We can use strong induction to prove this inequality. Let $P(n)$ be the statement $f(n) > \alpha ^{n-2}$. We want to show that $P(n)$ is true whenever $n$ is an integer greater than or equal to 3.
BASIS STEP: First, note that
$α < 2 = f(3), \ α^2 = (3 + \sqrt{5})/2 < 3 = f4$, so $P(3)$ and $P(4)$ are true.
INDUCTIVE STEP: Assume that $P(j)$ is true, namely, that $f(j) > α^{j−2}$, for all integers j with $3 ≤ j ≤ k$, where $k ≥ 4$. We must show that $P(k + 1)$ is true, that is, that $f(k+1) > α^{k−1}$. Because α is a solution of $x^2 − x − 1 = 0$ (as the quadratic formula shows), it follows that $α^2 = α + 1$. Therefore,*
$α^{k−1} = α^2 \cdot α^{k−3} = (α + 1)α^{k−3} = α \cdot α^{k−3 + 1} \cdot α^{k−3} = α^{k−2} + α^{k−3}$.
By the inductive hypothesis, because $k ≥ 4$, we have
$f(k−1) > α^{k−3}, \ f(k) > α^{k−2}$.
Therefore, it follows that
$f(k+1) = f(k) + f(k−1) > α^{k−2} + α^{k−3} = α^{k−1}$.
Hence, $P(k + 1)$ is true. This completes the proof.
I am confused on first part of the inductive step, namely:
$α^{k−1}$ = $α^2$ · $α^{k−3}$
I don't understand why $α^{k−1} = α^2 \cdot α^{k−3}$ and I don't understand why $α^2 = α + 1$ since this is a "solution of $x^2 − x − 1 = 0$ (as the quadratic formula shows)"
