Show there exists a solution in $\mathbb{N}$ to $x^4-y^4=2z^2$ with $\gcd(x,y,z)=1$

Question

I am allowed to assume that $x^4-y^4=2z^2$ already has one solution in $\mathbb{N}$. From this I have shown that there exists a solution in $\mathbb{N}$ with $x,y$ both odd.

I am now trying to show that there exists a solution with $\gcd(x,y,z)=1$. My guess is that if $x,y,z$ already have a common divisor, then I can find another solution by cancelling/dividing $x$ by something dependent on $d$, and similar for $y,z$.

I started by assuming that $d$ is a common divisor of $x,y,z$. That is $x=da,y=db,z=dc$. Then $d^4(a^4-b^4)=2d^2c^2 \implies d^2$ divides $2c^2$. Here is where I am stuck. It feels like the natural thing to do here is to show that $\gcd(2,c^2)=1$, so that in fact $d^2$ divides $c^2$, but I cannot. And even then I don't know where to go from that point.

Answer 1

As you stated, you've shown there exists a solution with $x$ and $y$ both odd. Thus, you have that the common factor of $d$ must be odd as well. This means when you have $d^2 \mid 2c^2$, that $\gcd(2,d^2) = 1$, so $d^2 \mid c^2 \implies d \mid c$ (e.g., this is explained at If $a^2$ divides $b^2$, then $a$ divides $b$). As such, for some integer $e \ge 1$, you have $c = ed$, which means $z = (ed)d = ed^2$. When you plug this into your equation you get

$$x^4 + y^4 = 2z^2 \implies a^4d^4 + b^4d^4 = 2e^2d^4 \tag{1}\label{eq1A}$$

You can divide both sides by $d^4$ to get

$$a^4 + b^4 = 2e^2 \tag{2}\label{eq2A}$$

If $\gcd(a,b) = f \gt 1$, then $f \mid 2e^2 \implies f \mid e^2$. However, this means at least one prime factor of $f$ divides $e$, so it would have divided $x$, $y$ and $z$ initially, but that's not possible since the gcd value of $d$ was divided out. As such, $f = 1$, so $a$, $b$ and $e$ are relatively prime to each other, i.e., $\gcd(a,b,e) = 1$.