On integer solutions of $2\sqrt{\sqrt{2x}+\sqrt{y}}=\sqrt{x}+\sqrt{2y}$ 
$\textsf{Background}$
From the double-angle formula $\cos2\alpha=2\cos^2\alpha-1$, we can get $$\cos15^\circ=\sqrt{\frac{1+\cos30^\circ}2}=\frac{\sqrt{2+\sqrt3}}2$$ but we also know that it is equivalent to $\dfrac{\sqrt6+\sqrt2}4$.

This is an example of an equality such that $$2\sqrt{\sqrt{2x}+\sqrt{y}}=\sqrt{x}+\sqrt{2y}$$ after some rearranging. We can write $y$ in terms of $x$ without much bother: $$y=2+\frac x2+2\sqrt{1+\sqrt{2x}}-\sqrt{2x(1+\sqrt{2x})}.$$ But when are $x$ and $y$ integers? Here is a plot of the curve: 
Some obvious solutions are $(0,0)$, $(0,4)$, $(2,3)$ and $(32,0)$. Of course, $x=2k$ for some integer $k$, leaving us with $$y=2+k+2\sqrt{1+2\sqrt k}-2\sqrt{k(1+2\sqrt k)}$$ Hence this boils down to finding $k$ such that $$(1-\sqrt k)\sqrt{1+2\sqrt k}$$ is an integer. Any advances on this?

The stated condition leads easily to
$$(\sqrt x + \sqrt{2y} - 2\sqrt2)^2 = 8 - 4\sqrt y.$$
Hence $0 \le y \le 4.$ Trying each value of $y$ in turn, one finds that the only solutions with $x$ also an integer are the four already stated.
