<s>[INST] Prove that $(n+\sqrt{n^2 -1})^k$ will always be of the form$ (t+\sqrt{t^2 -1})$ where $n$, $k$, $t$ are natural numbers Show that $(n+\sqrt{n^2 -1})^k$ will always be of the form$ (t+\sqrt{t^2 -1})$ where $n$, $k$, $t$ are natural numbers
 [/INST] I enjoyed working this out-cool question man. Here's a straightforward inductive proof:
I'm writing $c$ instead of $n$.
Inductively assume $(c + \sqrt{c^2 - 1})^k = a + b \sqrt{c^2 - 1}$, where $b^2(c^2 - 1) = a^2 - 1$ (the base case is clear). That is, not only can you write it as $t + \sqrt{t^2 - 1}$, but also $t^2 - 1$ is a square times $c^2 - 1$ (I guessed this by working out the first $k$'s for $c =2,3$).
Now to see $k \Rightarrow k+1$, we have $$(c + \sqrt{c^2 - 1})^{k+1} = (a + b \sqrt{c^2 - 1})(c + \sqrt{c^2 - 1}) $$$$= (ac + b(c^2 - 1)) + (a + cb)\sqrt{c^2 - 1}$$It suffices to check that $$(ac + b(c^2 - 1))^2 - 1 = (a + cb)^2(c^2 - 1)$$Expanding both sides gives $$c^2 a^2 + 2c(c^2 - 1)ab + (c^2 - 1)^2 b^2 - 1 = (c^2 - 1)a^2 + 2c(c^2 - 1)ab + c^2(c^2 - 1)b^2$$Cancelling the $ab$ terms, pulling the $a^2$ term to the LHS, and the $b^2$ terms to the RHS, we get $$a^2 - 1=b^2(c^2 - 1) $$our inductive hypothesis. 
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