<s>[INST] The equation $x^4+y^4=z^2$ has no integer solution The equation $$x^4+y^4=z^2$$ has no integer solution for $(x, y, z), x \cdot y \neq 0 , z >0$. 
We suppose that there is a solution $(x, y, z)$. 
We consider the set $$M=\{z \in \mathbb{N} | \exists x, y \in  \mathbb{Z}: x^4+y^4=z^2, x \cdot y \neq 0 \} \subseteq \mathbb{N}$$ 
Without loss of generality, we suppose that $(x, y, z)=1$. 
One of $x$ and $y$ must be even and the other one odd. 
We suppose that $x=2k+1, k \in \mathbb{Z}$ and $y=2l, l \in \mathbb{Z}$, 
$\Rightarrow z=2m+1>0, m \in \mathbb{Z}$. 
$$x^4+y^4=z^2 \Rightarrow y^4=z^2-x^4=(z-x^2)(z+x^2)$$ 
It stands that $gcd(z-x^2, z+x^2)=2$. 
To show that $gcd(z-x^2, z+x^2)=2$, is the following the only way?? 
Let $(z-x^2, z+x^2)=d>1$. Then it has a prime divisor, let $p$. 
$p \mid d , d \mid z-x^2 \Rightarrow p \mid z-x^2$ 
$p \mid d , d \mid z+x^2 \Rightarrow p \mid z+x^2$ 
So $p \mid 2x^2 \Rightarrow p \mid 2 \text{ OR } p \mid x$ 
and $p \mid 2z \Rightarrow p \mid 2 \text{ OR } p \mid z$ 
When $p \mid x \text{ AND } p \mid z$, we have that $p \mid y^4 \Rightarrow p \mid y$, so $p \mid (x, y, z)=1$, a contradiction. 
So, it should be $p \mid 2 \Rightarrow p=2$. 
Is this correct so far?? How can we continue to show that $d=2$ ?? 
$$$$ 
We have the following two cases 
$$(1): \begin{cases}
z-x^2=2a^4\\ 
z+x^2=8b^4
\end{cases} \text{ with } (a, b)=1, a \equiv 1 \pmod 2 , a>0$$ 
$$(2): \begin{cases}
z-x^2=8b^4\\ 
z+x^2=2a^4
\end{cases} \text{ with } (a, b)=1, a \equiv 1 \pmod 2 , a>0$$  
Why do we have these two cases??
 [/INST] Since $x,z$ are odd, certainly $z\pm x^2$ are even, hence the gcd is at least $2$. Any common divisor of $z+x^2$ and $z-x^2$ also divides their sum $2z$ and their difference $2x^2$ and their product $y^4$, i.e., $\gcd(z-x^2,z+x^2)\mid \gcd(2x^2,y^4,2z)\mid 2\gcd(x^2,y^4,z)\mid 2\gcd(x^4,y^4,z^4)=2\gcd(x,y,z)^4=2$
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