$\sqrt{x^2+y^3}$ and $\sqrt{x^3+y^2}$ are rational Are there infinitely many pairs of different positive rational numbers $x,y$ such that $\sqrt{x^2+y^3}$ and $\sqrt{x^3+y^2}$ are rational?
Consider such a pair. Then we have $x^2+y^3=a^2$ and $x^3+y^2=b^2$ for some rationals $a,b$. So $(a^2-y^3)^3=x^6=(b^2-y^2)^2$. Does this lead to something useful?

Yes, there are infinitely many solutions in integers.
In particular, we have a solution with $y = 2x$ whenever $x + 4$ and $8 x + 1$ are both squares (i.e. $x^2 + (2x)^3 = x^2 (1 + 8 x)$ and $(2x)^2 + x^3 = x^2 (4 + x)$).  If $x + 4 = u^2$ and $8 x + 1 = v^2$, eliminating $x$ gives us the 
equation $8 u^2 - 31 = v^2$, which is similar to a Pell equation.  One solution
is $u = 2, v = 1$, and whenever $\pmatrix{u\cr v}$ is a solution so is
$ \pmatrix{3  & 1\cr 8 & 3\cr} \pmatrix{u\cr v}$.  This gives the sequence
$$ \pmatrix{u\cr v\cr} = \pmatrix{2\cr 1\cr}, \pmatrix{7\cr 19\cr}, \pmatrix{40\cr 113\cr}, \ldots $$
corresponding to 
$$ \pmatrix{x\cr y\cr} = \pmatrix{0\cr 0\cr}, \pmatrix{45\cr 90\cr}, \pmatrix{1596\cr 3192\cr}, \ldots $$
