Integer solutions for $x^4 + 4xy^3 = z^2$

Question

Find all triplets $(x,y,z)$ of integers so that $$ x^4 + 4xy^3 = z^2. $$

What I've done:

Suppose $x=0$. Then we see $z=0$ and hence $(0,y,0)$ is a solution. Suppose $y=0$. Then we see $x^4 = z^2$ and hence $(x,0,\pm x^2)$ is a solution. Suppose $z=0$. Then we see $x^3=-4y^3$, which gives $x=y=0$ because otherwise the number of factors of $2$ on both sides cannot be equal.

We may now assume $x>0$ and $z>0$. I reasoned, if all three are even, $4$ must divide $z$, hence we may divide both sides by $16$, yielding a contradiction by infinite descent. Rewriting the equation as $4xy^3 = (z+x^2)(z-x^2)$ hasn't resulted in any good. Mathematica has found no solutions for $0<x<1000$ and $0<|y|<500$, so I think there are no more solutions. I haven't come really far yet, so any help is appreciated. Thanks.

Answer 1

I found an answer based on what MXYMXY posted earlier.

The crucial step is to observe that: $$ (2x^2)(z-x^2)(z+x^2) = 2x^2(z^2-x^4) = 2x^2(4xy^3) = (2xy)^3. $$ Note that from my analysis above we may conclude that all factors on the LHS are non-zero. Also note that $$(2x^2)+(z-x^2) = z+x^2.$$ Let $d = \gcd(2x^2,z-x^2) = \gcd(z-x^2,z+x^2) = \gcd(z+x^2,2x^2)$ and observe that $d^3|(2xy)^3$, thus $d|2xy$. Now set $a = \frac{2x^2}{d}$, $b = \frac{z-x^2}{d}$, $c = \frac{z+x^2}{d}$ and $n =\frac{2xy}{d}$. Note that $a$, $b$ and $c$ are pairwise coprime. It follows that: $$ abc = n^3 \quad \textrm{and} \quad a+b = c, $$ from which we conclude that $ab(a+b) = n^3$. Since $a$, $b$ and $a+b = c$ are pairwise coprime, we see that all three must be third powers, which is impossible by Fermat's Last Theorem.

This shows there are no further solutions.

Answer 2

If $a$, $b$ and $c$ are non-zero rational numbers, then $\frac{a}{b}+\frac{b}{c}=\frac{c}{a}$ has no solutions.

$pf)$ Let $\frac{a}{b}=x$, $\frac{b}{c}=y$, $\frac{c}{a}=z$. Note $xyz=1$

This implies that we have to find solutions to $xy(x+y)=1$ with $x,y$ rationals. However, this was already proved here.

Note that if $b^4+4bc^3$ had a nonzero integer solution, this implies $-\frac{b^2}{2c}\pm\frac{1}{2c}\sqrt{b^4+4bc^3}$ is a rational number. But, as @vrugtehagel pointed out in the MSE link above, this implies that there are rational solutions to $\frac{a}{b}+\frac{b}{c}=\frac{c}{a}$ if we set $a= -\frac{b^2}{2c}\pm\frac{1}{2c}\sqrt{b^4+4bc^3}$. A contradiction.

Thus, there are no nonzero solutions.

(In the link I provided, @vrugtehagel asked for integer solutions. But this can be extended to rational numbers as well.)