<s>[INST] 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.
 [/INST] 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.
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