How to show that the integral part of a binomial expansion is even?

Question

Question:

Show that the integral part of $(7+5\sqrt2)^{2n+1}$ is an even integer.

I can solve questions with power 'n', but have problem with powers such as'2n+1'. Is there any special condition?

Answer 1

Let $a_n=(7+5\sqrt2)^n+(7-5\sqrt2)^n$. Then $$ a_0=2, \quad a_1=14 \quad a_{n}=14a_{n-1}+a_{n-2} \mbox{ for } n \ge 2 $$ In particular, $a_n$ is always an even integer.

Because $-1 < 7-5\sqrt2 < 0$, the integer part of $(7+5\sqrt2)^n$ is $a_n-1$ when $n$ is even and $a_n$ when $n$ is odd. Hence the claim.