Showing $(x,y)$ pairs exist for $\sqrt{\quad\mathstrut}$ If we were to show that there exists infinitely many $(x,y)$ pairs in $\mathbb{Q}^2$ for which both $\sqrt{x^2+y^4}$ and $\sqrt{x^4+y^2}$ are rational. If the power root for $x$ and $y$ vary but never the same, how can we prove that we can always represent the outcome as a rational number? One way to do it would be to use the Pell equation. Is there any other ways?
 A: Let $x=y=t^2-1$. Then $x^2+y^3=x^3+y^2=t^2(t^2-1)^2$ so both squareroots are $t(t^2-1)$, which is rational provided $t$ is rational. This is infinitely many pairs $(x,y).$
With a little work and Pell's equation, one can find pairs $(x,y)$ of form $(2t,t)$ which work, if it's a concern that $x=y$ in the above example.
EDIT I just noticed the OP needs distinct $x,y$. With $x=2t$ and $y=t$ we have
$$x^2+y^3=t^2(t+4),\ \ x^3+y^2=t^2(8t+1).$$
Thus we want $t+4=a^2, 8t+1=b^2.$ Solving the first and putting it into the second gives the Pell type equation
$$[1]\ \ b^2-8a^2=-31.$$
This has a solution $(b,a)=(13,5),$ while the associated Pell equation $b^2-8a^2=1$ has a solution $(b,a)=(3,1)$. So in the usual Pell fashion, infinitely many solutions to [1] may be obtained by multiplying out the expression
$$(13+5\sqrt{8})\cdot (3+\sqrt{8})^n$$
and putting it in the form $b+a\sqrt{8}$. Then such $a,b$ must be substituted back into one of $t+4=a^2,\ 8t+1=b^2$ to get the $t$ for which $(x,y)=(2t,t)$ gives rational squareroots.
