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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?

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What have you tried? We will be better able to help you if you provide where you got stuck. :) –  apnorton Mar 6 '13 at 23:46
If this is connected to a number-theory course you’re taking, it would also help if you said what point you’re to in the course. –  Lubin Mar 7 '13 at 0:04
Please, try to make the title of your question more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. –  Martin Sleziak Mar 16 '13 at 6:31
This post should have been a new question, since previously it concerned making $x^2+y^3$ and $x^3+y^2$ both squares, whereas in its now rewritten form it asks for $x^2+y^4$ and $x^4+y^2$ to be squares, an entirely different question. –  coffeemath Mar 21 '13 at 21:40

1 Answer 1

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.

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