Show that if $2+2\sqrt{28n^2+1}$ is an integer then it must be perfect square. As written in title, I want to prove that 

If $n$ is an integer, show that if $2+2\sqrt{28n^2+1}$ is an integer than it must be perfect square. 

I m struggling in making a start . Please help.

Here's another answer which uses the Pell equation.
As mentioned above, in order for $2 + 2\sqrt{28y^2+1}$ to be an integer, it is necessary and sufficient that $28y^2 + 1$ is a perfect square.
Hence we must look at the solutions to Pell's equation $x^2 - 28y^2 = 1$.
All solutions to this equation are of the form $x_n+y_n\sqrt{28} = \pm\left(127+24\sqrt{28}\right)^n, n \in \mathbb{Z}$.
This means $x_n = \pm\frac{1}{2}\left[\left(127+24\sqrt{28}\right)^n+\left(127-24\sqrt{28}\right)^n\right]$.
Also note that $2 + 2\sqrt{28y_n^2 + 1} = 2+2\sqrt{x_n^2} = 2+2|x_n| = 2 + \left(127+24\sqrt{28}\right)^n + \left(127-24\sqrt{28}\right)^n$.
But $127 + 24\sqrt{28} = \left(8+3\sqrt{7}\right)^2$ and $\left(8+3\sqrt{7}\right)\left(8-3\sqrt{7}\right) = 1$ so we finally obtain:
$$2+2\sqrt{28y_n^2+1} = \left(8+3\sqrt{7}\right)^{2n} + 2\left(8+3\sqrt{7}\right)^n\left(8-3\sqrt{7}\right)^n+\left(8-3\sqrt{7}\right)^{2n} = \left[\left(8+3\sqrt{7}\right)^n+\left(8-3\sqrt{7}\right)^n\right]^2$$
which is the square of an integer (you can check with the binomial theorem).
