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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.

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    $\begingroup$ It might be easier to use the recursion, rather than the irrational numbers. I think it is $A_{k+1}=4A_k-A_{k-1}$. Look at the remainders $\pmod{4n+3}$, find values of $k$ when $A_k=0\pmod{4n+3}$. There may be a pattern. $\endgroup$ – Empy2 Dec 20 '14 at 17:14
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    $\begingroup$ @Michael: I have tried that recurrence. There is no patterns. $\endgroup$ – DER Dec 20 '14 at 17:16
  • $\begingroup$ Hint: $(-\sqrt{3}+2)=(\sqrt{3}+2)^{-1}$ $\endgroup$ – Mihail Dec 20 '14 at 17:24
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    $\begingroup$ @DER: I bet it is extremely difficult to find such $n$s because even assuming that $4n+3$ is a prime for which $3$ is a quadratic residue, the divisibility condition depends on the Legendre symbol for $2+\sqrt{3}$. So, it is extremely likely that an infinite number of such $n$s exist (by searching them among suitable primes), but to compute every suitable $n$ looks really, really hard. $\endgroup$ – Jack D'Aurizio Dec 20 '14 at 17:36
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    $\begingroup$ @DER: $$\left(\frac{\xi}{p}\right)=\xi^{\frac{p-1}{2}}.$$ I assumed $p=4n+3$ and $\left(\frac{3}{p}\right)=1$, so that $\xi=2+\sqrt{3}\in\mathbb{F}_p$. Notice that: $$\xi^{\frac{p-1}{4}}+\xi^{\frac{1-p}{4}}\equiv 0\pmod{p}$$ is equivalent to: $$\xi^{\frac{p-1}{2}}\equiv -1\pmod{p}.$$ $\endgroup$ – Jack D'Aurizio Dec 20 '14 at 17:39
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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 $$2^{n+2}\sum_k{n+1\choose2k}(3n+3)^k\\=2^{n+2}\sum_k{n+1\choose2k}(-n)^k\pmod{4n+3}\\ =2^{n+2}\Re(1+i\sqrt{n})^{n+1}$$

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    $\begingroup$ These terms are $0\pmod{4n+3}$ if the original numerators are $0\pmod{4n+3}$. I have added and subtracted multiples of $4n+3$. I don't think this is the answer, but hoped it might lead somewhere. $\endgroup$ – Empy2 Dec 22 '14 at 6:06

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