<s>[INST] $X^4 - 4Y^4 = -Z^2$ has no solutions in non zero integers I am trying to prove that $X^4 - 4Y^4 = -Z^2$ has no solutions in non zero integers. I know there are similar questions on MS, but that minus signs before the $Z$ gives me a hard time.
For the moment, I converted the equation to the equivalent $X^4 + Z^2 = 4Y^2$, and I supposed that $(x,y,z)$ is a primitive solution for this equation. Thus we have $(x^2)^2 +z^2 = (2y^2)^2$.
Now $(2y^2)^2$ is even, it tells us that $x$ and $z$ must have the same parity. But we know by Diophante's equation that there exist $a$ and $b$ with $a > b$, $(a, b) = 1$. If $a \not \equiv b \pmod 2$, we have $x^2 = a^2 - b^2, z = 2ab$ and $2y^2 = a^b + b^2$. But this is impossible, because of $2y^2 = a^2 + b^2$, $a$ and $b$ should have the same parity, which is not our hypothesis.
If $a$ and $b$ have the same parity, they are both odd (otherwise they are not coprime) and we have $x^2 = \frac{a^2 - b^2}{2}$, $z = ab$ and $2y^2 = \frac{a^2 + b^2}{2}$. But once again, as $(a, b) = 1$ and are both odd, they have no factor "2" in common thus $2y^2 = \frac{a^2 + b^2}{2}$ is impossible.
Therefore the equation $X^4 - 4Y^4 = -Z^2$ has no solution in $\mathbb{N} \setminus \{0\}$.
What do you think of this argument? My first idea was to use the infinite descent, but in this case I did not manage to do it.
My second idea was to prove that it was equivalent to another equation such as $X^4 + Y^4 = Z^2$ which we know has no solutions such that $XYZ \neq 0$ but I did not have more success.
 [/INST] For another way of showing your required conclusion, I give you the following:

Hint: 
  $$Z^2 = 4Y^4 - X^4 = (2Y^2 + X^2)(2Y^2 - X^2)$$

Can you take it from here?
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