The equation $x^4+y^4=z^2$ has no integer solution

Question

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??

Answer 1

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., $\gcd(z-x^2,z+x^2)\mid \gcd(2x^2,y^4,2z)\mid 2\gcd(x^2,y^4,z)\mid 2\gcd(x^4,y^4,z^4)=2\gcd(x,y,z)^4=2$

Answer 2

Since the discussion in the comments is getting complicated, let's do this directly.

Let $d=\gcd(z-x^2, z+x^2)$. Clearly $2|d$ since $2|z-x^2$ and $2|z+x^2$, seeing as how each of $z,x^2$ are odd, so we may write $d=2k$.

Now, for any divisor $f|(z-x^2)$ and $f|(z+x^2)$ then $f$ divides sums and differences of these things, so for our gcd we have

$$d|(z-x^2)+(z+x^2)=2z\qquad d|(z+x^2)-(z-x^2)=2x^2$$

Then as $d=2k$ we see that $k|x^2$ and $k|z$. Because $\gcd(x^2,z)=1$, there are integers $u,v$ such that $ux^2+vz=1$, hence $k|(ux^2+vz)\implies k|1$, i.e. $k=\pm 1$. We conclude $d=2$.