Show that if $2+2\sqrt{28n^2+1}$ is an integer then it must be perfect square.

With suggestions of @barak manos and @ Alqatrkapa I Noted that I can improve my post.

Notice that $2+2\sqrt{28n^2+1}$ is an even integer. Also, $28n^2+1$ is a perfect square of an odd integer say $m$ (Because $28n^2+1$ is odd itself).

Now, $$28n^2=m^2-1=(m+1)(m-1)\implies 7n^2=(\frac{m+1}{2})(\frac{m-1}{2})$$

Hence, $(\frac{m+1}{2})=7a^2$,$(\frac{m-1}{2})=b^2$ or $(\frac{m+1}{2})=b^2$, $(\frac{m-1}{2})=7a^2$. This is because $7n^2$ is $7$ times of a square and thus the right side is also $7$ times a perfect square. This is only possible when one of them is $7$ times of a square and other is simply a square as $(Square * Square=Square)$, you can say that there is possibility that they both are not squares but the product is (like $2*8=16=4^2$), but notice that $(\frac{m+1}{2})$ and $(\frac{m-1}{2})$ are consecutive integers and hence coprime

If $\frac{m+1}{2}=7a^2$ and $\frac{m-1}{2}=b^2$ then $b^2\equiv-1\mod (7)$, a contradiction.

Hence, $\frac{m-1}{2}=7a^2$ and $\frac{m+1}{2}=b^2$. Hence, $2+2m=4b^2$ a perfect square.

This stuff comes from a Pell equation.

It seems likely that https://en.wikipedia.org/wiki/John_Pell was in France on many occasions, in which case he may well have said

Au fait, je m'appelle Pell.

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For the curious, the values of $n$ that work are $$ 0, \; \; 24, \; \; 6096, \; \; 1548360, \; \; 393277344, \; \; \cdots $$ satisfying $$ n_{j+2} = 254 n_{j+1} - n_j $$ and $w = \sqrt{2 + 2 \sqrt{28 n^2 + 1}}$ $$ 2, \; \; 16, \; \; 254, \; \; 4048, \; \; 64514, \; \; \cdots $$ satisfying $$ w_{j+2} = 16 w_{j+1} - w_j $$

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).