<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] There are not solution. In fact we have
$$y=(\sqrt p-\sqrt x)^4=p^2+6px+x^2-4(p+x)\sqrt{px}$$ One has three possible ways:
►If $x=p$ then $y=8p^2-8p^2=0$
►If $p+x=0$ then $y=2p^2-6p^2\lt0$
►If $x=pz^2$ then $y=p^2(z-1)^4$ which implies the given equation becomes
$(z+z-1)\sqrt p=\sqrt p\Rightarrow z=1\Rightarrow x=p$ which is the first possibility.
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