Baltic Way 2019

Let's denote $$\sqrt{12n^2+1}=2k+1(\text{Eq-6})$$ because if the number $\sqrt{12n^2+1}$ is integer then it is odd. We have to prove that $$\sqrt{\frac{\sqrt{12n^2+1}+1}{2}}=\sqrt{\frac{(2k+1)+1}{2}}=\sqrt{k+1}$$ is an integer, that is, that $k+1$ is a perfect square.

By squaring both sides of (Eq-6) we get that $12n^2+1=4k^2+4k+1$ or $k(k+1)=3n^2$. As $k$ and $k+1$ are coprime then we have two options: $k=3a^2$ and $k+1=b^2$ or $k=a^2$ and $k+1=3b^2$ for some positive integers $a$ and $b$. But the second option is not possible as that would mean that $a^2=3b^2-1=2(\mathrm{mod}3)$.