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Prove that if

$$k = 2 + 2\sqrt{12n^2 + 1}$$

is an integer then it is a square.

Can anyone help me with this? All I know is that k is an integer if and only if ${12n^2 + 1}$ is a square. What do I do next?

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  • $\begingroup$ The general $n$ that satisfies this is $$n=\frac{(7+2\sqrt{12})^m+(7-2\sqrt{12})^m}{2}$$ by the general solution to the Pell equation. $\endgroup$ – Thomas Andrews Feb 1 '15 at 4:33
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let $$\sqrt{12n^2+1}=m\Longrightarrow 12n^2+1=m^2\Longrightarrow \dfrac{m-1}{2}\cdot\dfrac{m+1}{2}=3n^2$$ because $m$ is odd numbers,so $\dfrac{m-1}{2},\dfrac{m+1}{2}\in N^{+}$

since $$\gcd\left(\dfrac{m+1}{2},\dfrac{m-1}{2}\right)=1$$

case 1:

$$\dfrac{m-1}{2}=3u^2,\dfrac{m+1}{2}=v^2,uv=n$$ $$\Longrightarrow 2+2\sqrt{12n^2+1}=2m+2=4v^2$$

case 2: $$\dfrac{m-1}{2}=u^2,\dfrac{m+1}{2}=3v^2\Longrightarrow 3v^2=u^2+1$$ since $3v^2\equiv 0,3\pmod 4$,and $u^2+1\equiv 1,2\pmod 4$

so that's impossible

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$m^2 = 12 n^2 + 1$ implies $m^2 - 12 n^2 = 1$ so $$(m + n\sqrt{12} )= (7 + 2\sqrt{12} )^N$$ for some $N\ge 0$ (see also http://en.wikipedia.org/wiki/Pell%27s_equation, just mind you, their $n$ is our $12$).

We also have the conjugate equality: $$(m - n\sqrt{12} )= (7 - 2\sqrt{12} )^N$$ and , therefore, for $\sqrt{12n^2 + 1} = m$ we get, as @Thomas Andrews indicated

$$\sqrt{12n^2 + 1} = m = \frac{1}{2}(\, (7 + 2\sqrt{12} )^N + (7 - 2\sqrt{12} )^N) $$ and so

$$2 + 2\sqrt{12n^2 + 1} = (7 + 2\sqrt{12} )^N + (7 - 2\sqrt{12} )^N) + 2$$

Note however that $$7 \pm \sqrt{12} = (2 \pm \sqrt{3})^2 $$ and so we get

$$2 + 2\sqrt{12n^2 + 1} = (2 + \sqrt{3} )^{2N} + (2 - \sqrt{3} )^{2N}) + 2\cdot (2 + \sqrt{3} )^N \cdot (2 - \sqrt{3} )^N = \\ = (\,( 2 + \sqrt{3})^N + (2-\sqrt{3})^N)^2$$

and inside the bracket we have a natural number.

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