Prove that all positive integers $n$, $(1-{\sqrt 5})^n$ can be written in the form $a-b{\sqrt 5}$ where $a$ and $b$ are positive integers Prove, by induction, that all positive integers $n$, $(1-{\sqrt 5})^n$ can be written in the form $a-b{\sqrt 5}$  where $a$ and $b$ are positive integers.
I understand these idea of proof by induction and this was a different type of question that I'm used too and wasn't sure on how to approach it as I'm not entirely confident with proving things with induction yet.
Any help would be appreciated.
 A: 1-Base Step: $$For\ n=1,\ {{\left( 1-\sqrt{5} \right)}^{1}}=a-b\sqrt{5},\ with\ a=1\,and\ b=1$$
2-Inductive step: Assume that ${{\left( 1-\sqrt{5} \right)}^{n}}=a-b\sqrt{5},\ Consider\ {{\left( 1-\sqrt{5} \right)}^{n+1}}$
$\begin{align}
  & {{\left( 1-\sqrt{5} \right)}^{n+1}}=\left( 1-\sqrt{5} \right){{\left( 1-\sqrt{5} \right)}^{n}} \\ 
 & \quad \quad \quad \quad \ \ =\left( 1-\sqrt{5} \right)\left( a-\sqrt{5}b \right) \\ 
 & \quad \quad \quad \quad \ \ =a-a\sqrt{5}+5b-b\sqrt{5} \\ 
 & \quad \quad \quad \quad \ \ =\left( a+5b \right)-\left( a+b \right)\sqrt{5} \\ 
\end{align}$
So the inductive case holds. Now by induction we see that the assumption is true.
A: For a direct proof let use binomial theorem for $(1-{\sqrt 5})^n$ that is
$$(x+y)^n = {n \choose 0}x^n y^0 + {n \choose 1}x^{n-1}y^1 + {n \choose 2}x^{n-2}y^2 + \cdots + {n \choose n-1}x^1 y^{n-1} + {n \choose n}x^0 y^n$$
By induction let consider


*

*base case: for $n=1\implies (1-{\sqrt 5})^1=1-\sqrt 5$

*induction step: assume that $(1-{\sqrt 5})^n=a-b{\sqrt 5}$ is true then


$$(1-{\sqrt 5})^{n+1}=(1-{\sqrt 5})^n(1-{\sqrt 5})\stackrel{I.H.}=(a-b{\sqrt 5})(1-{\sqrt 5})=\\=a-a\sqrt 5-b\sqrt 5+5b=a+5b-(a+b)\sqrt 5$$
that is
$$(1-{\sqrt 5})^{n+1}=a+5b-(a+b)\sqrt 5$$
A: Hint: Note that $x^{n+1} = x^n\cdot x$.
A: Hint: $\;(1-\sqrt{5})(a-b\sqrt{5})=a+5b-(a+b)\sqrt{5}\,$.
