<s>[INST] Show that the equation $y^2 = x^3 +xz^4$ has no solutions in nonzero integers $x,y,z$. 
Show that the equation $y^2 = x^3 +xz^4$ has no solutions in nonzero integers $x,y,z$. 

There is given hint: Suppose that there is a solution. First show that it can be reduced to a solution satisfying $\gcd(x,z)=1$. Then use the fact that $x^3+ xz^4= x(x^2+z^4)$ is a perfect square to show that there are no solutions other than $x=y=0$.
The hint tells to prove by contradiction approach. So, $y\mid x,z\implies y\mid p$, where $p=(x,z)$. Hence, $\exists X,Z\in \mathbb{Z},\,x=Xp, z=Zp\implies y^2= X^3p^3 +XpZ^4p^4$$\implies y^2= Xp^3(X^2+Z^4p^2)$.

Is it valid to conclude from the last line, any of the below two separate conclusions : 
1. The set of common divisors of $y^2, x^3, xz^4$ are the same; hence let $gcd(y^2, x^3, xz^4)=p'$
2.$\,\,p^3\mid y^2\implies \exists y=p^2Y, \,\,\,\,p^2\mid y$
 [/INST] This is not much more than a lengthy remark, but it's worth posting as an answer, I think, because it shows that proving this is at least as hard as proving the strong version of Fermat's Last Theorem for $n=4$. That is, suppose that $a^4+b^4=c^2$ with $abc\not=0$. Then, letting $x=a^2$, $z=b$, and $y=ac$, we first have $x^2+z^4=c^2$, and thus $x(x^2+z^4)=xc^2=(ac)^2=y^2$ with $xyz=a^3cb\not=0$. So if $y^2=x^3+xz^4$ has no solutions with $xyz\not=0$, then $a^4+b^4=c^2$ has no solutions with $abc\not=0$ either.
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