We work on the ring of Gaussian integers $\mathbb{Z[\i]}$. Note that $2$ is not a factor of $z$ by mod 4 reasons. Write $(x^{2}+iy^{2})(x^{2}-iy^{2})=2z^{2}$. Let $d$ the gcd of the 2 factors at the LHS. Then $d\mid x^{2}$ and $d\mid iy^{2}$ (since $gcd(d,2)=1$). By taking the norms, and using that $gcd(x,y)=1$ we conclude that $d$ is a unit, hence the 2 factors are coprime. Hence: $x^{2}+iy^{2}=2(a+bi)^{2}$ $x^{2}-iy^{2}=(c+di)^{2}$ or vice versa, Hence, $x^{2}= 2a^{2}-2b^{2}$ and $y^{2}=2cdi$ and for obvious reasons $ad=-bc$ ($(a+bi)(c+di)\in \mathbb{R}$). The details are left for the patient reader.

oVlad wrote: Determine all triples of positive integers $(x,y,z)$ such that $x^4+y^4 =2z^2$ and $x$ and $y$ are relatively prime. Turn it into the form $(\frac{x^4-y^4}{2})^2=z^4-x^4y^4$,where we have the famous equation $a^4-b^4=c^2$ and get $x=y=z=1$.

how does a^4-b^4=c^2 make (a.b.c)=(1,1,1)

David_Kim_0202 wrote: how does a^4-b^4=c^2 make (a.b.c)=(1,1,1) In this problem,we have $gcd\left(a,b\right)=1$ if $c\ne 0$,let $\left(a_0,b_0,c_0\right)$ be a solution. then $a_0^2=m^2+n^2,b_0^2=m^2-n^2,c_0=2mn$ for some $m,n$. $\left(m,n,a_0b_0\right)$ is also a solution and the ”$a$“ is smaller. So we can always let $a$ be smaller,contradiction! so $c=0,a=b=1$

a-ha thanks so this can be solved by pitagoras pairs !!