Let $k,n$ be positive integers.
Let ${\displaystyle a_{n,k}=\frac{(3+2 \sqrt{2 k})^{n}-(3-2 \sqrt{2 k})^{n}}{\sqrt{2k}}}$.
$a_{n,k}$ is an integer (equal to ${\displaystyle2\sum_{p=0}^{n/2-1}\binom{n}{2 p+1} 3^{n-2 p-1} 2^{3 p} k^{p}}$).
How could I show that the 2-adic valuation of $a_{n,k}$ is equal to the one of $n$ plus 2?
$\nu_2(a_{n,k})=\nu_2(n)+2$.
Try binomial expansion of $(3 \pm 2 \sqrt{2k}))^n$. When you take the difference between them and divide by $\sqrt{2k}$ (which will give an integer for all integers $n$) you should find that the surviving terms will have different $2$-adic norms and all you need to do is pick the term with the largest norm.
Let's take $n=3$ as an example. Expansion gives
$(3\pm2\sqrt{2k})^3=27\pm54\sqrt{2k}+72k\pm16k\sqrt{2k}$
Taking the difference and dividing by $\sqrt{2k}$ leads to
$108+32k$
where $108$ has two powers of $2$ thus $|108|=1/4$, and $32k$ has at least five powers of $2$ hence $|32k|\le1/32$. The former term is larger in absolute value than all others, so the sums must have $\nu=2,|·|=1/4$ as claimed for odd $n$.
Try it for general $n$ and see what happens.