I was solving the following question a while ago:
Let $\alpha$ and $\beta$ be the roots of $x^2 - 6x +1 = 0$. Show that $\alpha^n + \beta^n$ is an integer for any integer $n\ge0$ and it is not divisible by 5.
I proved it with Mathematical Induction in the following manner:
Proof:
Base case : For $n = 1$, $\alpha^1 + \beta^1 = 6$ {I skipped $n = 0$ for a reason}
Induction: Let $n= k$, where $k\in Z$ and $k \ge 0$
Now, we have to prove that $\alpha^{k+1} + \beta^{k+1}$ is also an integer.
Proof: $(\alpha + \beta)(\alpha^{k} + \beta^{k}) \in Z$ {Given}
$=>(\alpha + \beta)(\alpha^{k} + \beta^{k}) = \alpha^{k+1} + \beta^{k+1} + \alpha\beta(\alpha^{k-1} + \beta^{k-1}) $.......(1)
This is where I got confused, Clearly $\alpha^0 + \beta^0$ and $\alpha^1 + \beta^1$ are integers. So, we should be able to claim that $(\alpha^{k-1} + \beta^{k-1})$ is also a integer. If we assume that $k =1$ then $k-1 = 0$, so I think induction should work this way too.
If $(\alpha^{k-1} + \beta^{k-1})$ is also a integer, then its easy to see that $\alpha^{k+1} + \beta^{k+1}$ is also an integer. Hence, we are done.
So, Is it fair to assume that $(\alpha^{k-1} + \beta^{k-1})$ is an integer ? {I am intentionally skipping the second part of the given question because its not the main motive.}Also, I learned about Mathematical Induction recently.
