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


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, $$(3+\sqrt5+3-\sqrt5)\cdot((3+\sqrt5)^k+(3-\sqrt5)^k)=(3+\sqrt5)^{k+1}+(3-\sqrt5)^{k+1}+3((3+\sqrt5)^k+(3-\sqrt5)^k)+\sqrt5(3+\sqrt5)^k-\sqrt5(3-\sqrt5)^k$$
