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
Let $\sqrt {n^2-1}=\alpha$

Then $(n+\alpha)(n-\alpha)=1$ and it is easy to see (prove by induction) that $(n+\alpha)^k = t+u\alpha \text{ and } (n-\alpha)^k = t-u\alpha$ with $t$ and $u$ integers so: $$1=(n+\alpha)^k(n-\alpha)^k = (t+u\alpha)(t-u\alpha)= t^2-u^2\alpha^2 $$

So $$u\alpha = \sqrt{t^2-1} \text{ and } (n+\alpha)^k=t+\sqrt{t^2-1}$$
Let, $$f(k)=(2+\sqrt[]{2^2-1})^k$$ -------(1)

consider $k=1$

$$f(1)=(2+\sqrt[]{2^2-1})^1$$

i.e. $$f(1)=(2+\sqrt[]{2^2-1})=\frac{1}{2-\sqrt[]{2^2-1}}$$

consider $k=2$

$$f(2)=(2+\sqrt[]{2^2-1})^2$$

i.e. $$f(2)=(4+(2^2-1)+4\sqrt[]{2^2-1})$$

i.e. $$f(2)=(7+4\sqrt[]{2^2-1})$$

i.e. $$f(2)=(7+\sqrt[]{(4^2)(2^2-1)})$$

i.e. $$f(2)=(7+\sqrt[]{7^2-1})=\frac{1}{7-\sqrt[]{7^2-1}}$$
Assume $f(n)$ holds for $k=n$

i.e. $$f(n)=(2+\sqrt[]{2^2-1})^n=(t+\sqrt[]{t^2-1})=(a+b\sqrt[]{3})$$ -------(2)

where a and b are natural numbers

i.e. $$f(n)=(2+\sqrt[]{2^2-1})^n=(t+\sqrt[]{t^2-1})=(a+\sqrt[]{3b^2})$$ 

for the assumption to hold, $$3b^2 = a^2-1$$ -------(3) 

consider $k=n+1$

i.e.
$$f(n+1)=(2+\sqrt[]{2^2-1})^{n+1}=(t+\sqrt[]{t^2-1})(2+\sqrt[]{2^2-1})$$

we also have, $$f(n+1) =(a+b\sqrt[]{3})(2+\sqrt[]{3})$$

where a and b are natural numbers

i.e. $$f(n+1)=(2a+3b)+(a+2b)\sqrt[]{3}$$ -------(4)

i.e. $$f(n+1)=(2a+3b)+\sqrt[]{3(a+2b)^2}$$

i.e. $$f(n+1)=(2a+3b)+\sqrt[]{3(a^2+4b^2+4ab)}$$

i.e. $$f(n+1)=(2a+3b)+\sqrt[]{3a^2+12b^2+12ab)}$$

i.e. $$f(n+1)=(2a+3b)+\sqrt[]{3a^2+12b^2+12ab + (a^2-a^2) + (3b^2-3b^2)}$$ -------(5)

i.e. $$f(n+1)=(2a+3b)+\sqrt[]{(3a^2+12b^2+12ab +a^2) -a^2 + (3b^2-3b^2)}$$

i.e. $$f(n+1)=(2a+3b)+\sqrt[]{(4a^2+12b^2+12ab) + (3b^2-3b^2) -a^2}$$

i.e. $$f(n+1)=(2a+3b)+\sqrt[]{(4a^2+12b^2+12ab) + (-3b^2+3b^2) -a^2}$$

i.e. $$f(n+1)=(2a+3b)+\sqrt[]{(4a^2+12b^2+12ab -3b^2) +3b^2 -a^2}$$

i.e. $$f(n+1)=(2a+3b)+\sqrt[]{(4a^2+9b^2+12ab) + (3b^2 -a^2)}$$

i.e. $$f(n+1)=(2a+3b)+\sqrt[]{(2a+3b)^2 + (3b^2 -a^2)}$$ -------(6)

by careful observation, using expressions (3) and (6) we have $$f(n+1)=(2a+3b)+\sqrt[]{(2a+3b)^2 + (a^2 -1 -a^2)}$$

i.e. $$f(n+1)=(2a+3b)+\sqrt[]{(2a+3b)^2 -1 + (a^2-a^2)}$$

i.e. $$f(n+1)=(2a+3b)+\sqrt[]{(2a+3b)^2 -1}$$ which is indeed of the form $$f(n+1)=(t+\sqrt[]{t^2-1})$$ where $t=(2a+3b)$