Diophantine equation $x^4+5y^4=z^4$ I am trying to find all positive integer solution $(x,y,z)$ of equation $x^4+5y^4=z^4$. 
Here I fould: $(x,y,z)=(1,2,3)$ and $(d,2d,3d)$. I try to prove if $(x,y)=1$ then $(1,2,3)$ is the unique solution of equation. Could anyone help me for this question? 
 A: I experimentally searched the possibility of more principal solutions for this equation.
$x^4 +5y^4=z^4$
$z^4≡ x^4\mod y^4$⇒ $x^4<y^4$ or $x<y$
$z^4≡ x^4\mod 5$ ⇒ $x^4 < 5$; the only possible number for x is $1$, that is $x=1$⇒ $(y=2)$ and $z=3$ are the principal solutions. 
Also we may write:
$(z^2+x^2)(z+x)(z-x)=5y^4$
Following cases can be considered:
a: $z-x =2$, that is z and x are two consecutive odd or even numbers. One solution is already found i.e. $(z=3)$ and $(x =1)$ and $y=2$. I checked by Python up to $z=10^6$ and got no more solutions.
b: $z-x=1$, that is x and z are two consecutive numbers. This can not be acceptable because $z>y>x$ i.e. the next consecutive number to x can only be y.
c: $z-x=5$, and $(z^2+x^2)(z+x)=y^4$; I checked this up to $z=10^6$ and I got no result.
c:$x+z=5$ that means:we have :$(x=1, z=4), (x=2, z=3), (x=3, z=2) and finally(x=4, z=1)$ which non of them give integer solution for y.
d:$z^2+x^2=5$ , the possible values are $(x=1)$ and$( z=2)$ which do not give an integer solution for y.
e: $z^2-x^2=5$ or $z^2+x^2=y^4$
I checked it up to $ z=10^6$ and could not found any integer solution for y.
Conclusion: equation $x^4 +5y^4=z^4$ is one case of general equation of form $x^4 +Dy^4=z^4$ in which $D=5$ . Lagrange showed that this type of equation depending on certain values of D can have integer solutions in limited numbers.The equation $x^4 +5y^4=z^4$ is one of these equations that have only one set of principal solutions $(x=1, y=2, z=3)$.
