parmenides51 wrote: a) Prove that for each positive integer $n$ there is a unique positive integer $a_n$ such that $$(1 + \sqrt5)^n =\sqrt{a_n} + \sqrt{a_n+4^n} .$$ Equation $(1+\sqrt 5)^n=\sqrt x+\sqrt {x+4^n}$ implies $(1+\sqrt 5)^n-\sqrt x=\sqrt {x+4^n}$ And so (squaring) $(1+\sqrt 5)^{2n}-2\sqrt x(1+\sqrt 5)^n+x=x+4^n$ And so $\sqrt x=\frac{(1+\sqrt 5)^{2n}-4^n}{2(1+\sqrt 5)^n)}$ $=\frac{(\sqrt 5+1)^n-(\sqrt 5-1)^n}2$ And so $a_n=b_n^2$ where $b_n=\frac{(\sqrt 5+1)^n-(\sqrt 5-1)^n}2$ Looking at $b_n$, we get $b_0=0$, $b_1=1$ and $b_{n+2}=2\sqrt 5b_{n+1}-4b_n$ A simpl;e induction proves that $b_{2k+1}$ is an integer and that $b_{2k}$ is the product of $\sqrt 5$ by an integer. And so $a_n=b_n^2$ always is a positive integer. Q.E.D.

Interesting If we change 5 by p a prime does exist a unique positive integer y such that $(1+\sqrt{p})^n = \sqrt{y} + \sqrt {y + (p-1)^n}$ ?

Moubinool wrote: Interesting If we change 5 by p a prime does exist a unique positive integer y such that $(1+\sqrt{p})^n = \sqrt{y} + \sqrt {y + (p-1)^n}$ ? Exactly the same computation gives $a_n=b_n^2$ where $b_n=\frac{(\sqrt p+1)^n-(\sqrt p-1)^n}2$ Looking again at $b_n$, we get $b_0=0$, $b_1=1$ and $b_{n+2}=2\sqrt pb_{n+1}-(p-1)b_n$ The same simple induction proves that $b_{2k+1}$ is an integer and that $b_{2k}$ is the product of $\sqrt p$ by an integer. And so $b_n^2$ always is an integer. Q.E.D.