<s>[INST] Mathematical Induction: Can I assume that $P(k-1)$ is also true in the induction hypothesis? Suppose that $y \in \mathbb{R}$, $y \neq 0$ and $y + \frac{1}{y} \in \mathbb{Z}$. Prove using mathematical induction that $y^{n} + \frac{1}{y^{n}}$ is an integer for all $n \geq 1$.
Basis step: The statement is true for $n=1$ since: $y^{1} + \frac{1}{y^{1}} = y^{n} + \frac{1}{y^{n}} \in \mathbb{Z}$.
Induction hypothesis: Assume that $y^{k} + \frac{1}{y^{k}}$ and $y^{k-1} + \frac{1}{y^{k-1}}$ are integers for all $n \geq 1$.
Induction step: We show that $y^{k+1} + \frac{1}{y^{k+1}}$ is true.
$y^{k+1} + \frac{1}{y^{k+1}} = (y^{k} + \frac{1}{y^{k}})(y + \frac{1}{y}) - (\frac{y^{k}}{y} + \frac{y}{y^{k}})$
$y^{k+1} + \frac{1}{y^{k+1}} = (y^{k} + \frac{1}{y^{k}})(y + \frac{1}{y}) - (y^{k-1} + \frac{1}{y^{k-1}})$
Since it is given that $y + \frac{1}{y} \in \mathbb{Z}$ and by the induction hypothesis, $y^{k} + \frac{1}{y^{k}}$, $y^{k-1} + \frac{1}{y^{k-1}} \in \mathbb{Z}$:
$\Rightarrow (\mathbb{Z})(\mathbb{Z})-\mathbb{Z}=\mathbb{Z}$ (Closure property for integers)
Therefore, $y^{n} + \frac{1}{y^{n}}$ is an integer for all $n \geq 1$.
This is the only way I was able to do the proof. However, I want to know if I am allowed to assume that $P(k-1)$ is also true in the induction hypothesis since as far as I know we can only assume that $P(k)$ is true. Would also appreciate feedback and alternative solutions for my proof!
 [/INST] For $n=2$ and $n=1$ the statement is correct.
Because, $$\left(y^2+\frac{1}{y^2}\right)=\left(y+\frac 1y\right)^2-2$$
Then ,suppose that $n=k$ and $n=k-1$ are also correct.....
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