<s>[INST] Solve for $\sqrt{x}+\sqrt[4]{y}=\sqrt{p}$ with $x$, $y$ are positive integers and $p$ is a prime number For that given problem, I have constructed a solution to it:

Squaring both sides, we get $x+\sqrt{y}+2\sqrt[4]{x^2y}=p$. Suppose $\sqrt{y}$ and $\sqrt[4]{x^2y}$ are integers, we can let $y=y'^2$, $y'$ being a positive integer. Subbing to the equation, we get $x+y'+2\sqrt{xy'}=p$, so $(\sqrt{x}+\sqrt{y'})^2=p$. We can see that $\sqrt{x}+\sqrt{y'}$ is a positive integer, and $x$ and $y'$ are positive, so $\sqrt{x}+\sqrt{y'}>1$, so $(\sqrt{x}+\sqrt{y'})^2$ has at least 3 positive divisors, contradiction.

The only problem here is that I can't show that $\sqrt{y}$ and $\sqrt[4]{x^2y}$ are integers, because $\sqrt[4]{x^2y}=\sqrt{x\sqrt{y}}$, and I cannot guarantee that $x\sqrt{y}$ is an integer, so I can't show that they are both integers.
I would really appreciate it if I can get a hint on this problem, thank you a lot in advance.
 [/INST] See that $\sqrt[4]y = (\sqrt p - \sqrt x)$ so
$$ y = (\sqrt p -\sqrt x)^4 = p^{2} - 4 \,  p \sqrt{px} + 6 \, p x - 4x \, \sqrt{px}  + x^{2} =p^2 +6px - \sqrt{px}(4x+4p) +x^2 $$ So you can easily see that $\sqrt{xp}$ is an integer.
Now $\sqrt y = (\sqrt p -\sqrt x)^2 = p + x - 2\sqrt{px}$ so $\sqrt y$ is an integer.
Also $\sqrt{x}\sqrt[4]y = x - \sqrt{px}$ which is an integer.
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