Floor function parity problem

Question

Prove that for every natural k this expression is always odd $⌊(5+\sqrt{19})^k⌋=A^k$ Progress that I' ve done is: I noticed $9^k<A^k<(9.5)^k$ Also I tried an induction approach, I used Binomial theorem, I rewrote the expression as $⌊(9+(\sqrt{19}-4))^k⌋$. But so far none of this approaches lead me to anything. Could you give me some tips.

Answer 1

Hint: For all integers $k \ge 0$,

$$f(k) = (5 + \sqrt{19})^k + (5 - \sqrt{19})^k \tag{1}\label{eq1A}$$

is an integer. You can fairly easily prove this using something like the binomial theorem, where the pairing of terms also gives useful parity information. In addition, note $0 \lt 5 - \sqrt{19} \lt 1$.