if $S_n = (3 + \sqrt{5})^n + (3 - \sqrt {5})^n$ then show that $S_n$ is an integer by induction

Question

this is all that i have tried:

let $n=1$ so equation gives $(3 + \sqrt {5}) + (3 - \sqrt {5}) = 6$ which is an integer

so it is true for $n=1$

now let it be true for $k \ge n$ then we have:

$(3 + \sqrt {5})^k + (3 - \sqrt {5})^k$ is an integer

for k+1: $(3 + \sqrt {5})^k(3 + \sqrt {5}) + (3 - \sqrt {5})^k(3 - \sqrt {5})$

$= 3((3 + \sqrt {5})^k + (3 - \sqrt {5})^k) + \sqrt {5}((3 + \sqrt {5})^k - (3 - \sqrt {5})^k)$

$= (integer) + \sqrt {5}((3 + \sqrt {5})^k - (3 - \sqrt {5})^k)$

from here I started facing a problem. I wasn't able to prove $\sqrt {5}((3 + \sqrt {5})^k - (3 - \sqrt {5})^k)$ an integer.

can someone please help me out?

(ONLY USING INDUCTION!)

Answer 1

Use the fact that $$a^{n+1} + b^{n+1} = (a+b)(a^n + b^n) - ab(a^{n-1} + b^{n-1}),$$ so that if $a = 3 + \sqrt{5}$, $b = 3 - \sqrt{5}$, then $ab = 3^2 - 5 = 4$, hence $$S_{n+1} = 6S_n - 4S_{n-1}.$$ Consequently your induction base case should cover two consecutive values of $n$, and your induction step relies on the fact that the previous two values of $S$ are integers.

Answer 2

Let's do a double induction.

To show: $(3+\sqrt5)^k+(3-\sqrt5)^k$ and $\sqrt5(3+\sqrt5)^k-\sqrt5(3-\sqrt5)^k$ are integers.

Base Case: trivial

Inductive Case: You've shown why $(3+\sqrt5)^k+(3-\sqrt5)^k$ is an integer. And, $$(3+\sqrt5+3-\sqrt5)\cdot((3+\sqrt5)^k+(3-\sqrt5)^k)=(3+\sqrt5)^{k+1}+(3-\sqrt5)^{k+1}+3((3+\sqrt5)^k+(3-\sqrt5)^k)+\sqrt5(3+\sqrt5)^k-\sqrt5(3-\sqrt5)^k$$