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Problem:

Let $n \in \mathbb{N}$ such that $2+2\sqrt{12n^2+1}$ is the integer. Prove that $2+2\sqrt{12n^2+1}$ is perfect square.

I tried to found $n$ such that $\sqrt{12n^2+1}$ is integer, i.e. $12n^2 + 1 = k^2$. It is a Pell equation, and it has solution: $(n_0,k_0) = (0,1)$, $$\begin{cases} n_{i+1} &=& 7n_i + 2k_i \\ k_{i+1} &=& 24n_i + 7k_i.\end{cases}$$ You can see this link. But I don't know how the next step. Or maybe you have another approach.

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Let $$12n^2+1=4k^2+4k+1 \Rightarrow 3n^2=k^2+k=k(k+1)$$ This implies that $k=3a^2, k+1=b^2$ or $k=b^2, k+1=3a^2$. However, the second case is impossible as $$b^2 =3a^2-1 \equiv -1 \pmod 3$$ Thus, we have that $k+1$ is a perfect square.

Note $$2+2\sqrt{12n^2+1}=4k+4=(2b)^2$$ Our proof is done.

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  • $\begingroup$ Oh, thanks, the proof is very nice. $\endgroup$
    – GAVD
    Jun 16, 2016 at 9:38
  • $\begingroup$ It was used here that $\sqrt{a}$, where $a\ge 0$ is an integer, is either an integer or irrational (see this question) to claim that $12n^2+1$ must be a perfect square. $\endgroup$
    – user236182
    Jun 16, 2016 at 9:39
  • $\begingroup$ You were also assuming that $k\in\mathbb Z^+$. Otherwise there would also be the cases $k=-3a^2$, $k+1=-b^2$ and $k=-b^2$, $k+1=-3a^2$ or $k=0$ or $k+1=0$. $\endgroup$
    – user236182
    Jun 16, 2016 at 9:41

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