TITLE: Find the values of the positive integers $n$ such that: $\frac{(-\sqrt{3}+2)^{n+1}+(\sqrt{3}+2)^{n+1}}{4n+3}$ is positive integer
QUESTION [7 upvotes]: My question is as follow:
Find the values of the positive integers $n$ such that:
$$\frac{(-\sqrt{3}+2)^{n+1}+(\sqrt{3}+2)^{n+1}}{4n+3}$$ is positive integer.
I can see that for $n=1$ (among some other numbers) the condition is verified. However, I cannot go further with this procedure.

REPLY [2 votes]: FWIW, I have changed the problem; I don't know whether it helps.  It is a bit big to fit into a comment.
The numerator is $$2\sum_k{n+1\choose2k}2^{n+1-2k}3^k$$
Multiply each term by by $(4n+4)^k=1\pmod{4n+3}$ and get
$$2^{n+2}\sum_k{n+1\choose2k}(3n+3)^k\\=2^{n+2}\sum_k{n+1\choose2k}(-n)^k\pmod{4n+3}\\
=2^{n+2}\Re(1+i\sqrt{n})^{n+1}$$