Is it possible to reduce the diophantine equation $$x^4+4y^4=z^2,$$ to the diophantine equation $$x^4+y^4=z^2,$$ to show that there is no nontrivial solution?
Please assume the second equation is already known to have only trivial solutions.
I think Fermats descent can prove this, but I hope there is an easier way.
A nontrivial solution to the diophantine equation $$X^4+4Y^4=Z^2,\tag{1}$$ is a triplet $(x,y,z)\in\Bbb{Z}^3$ satisfying the equation with $xyz\neq0$. We will show that no nontrivial solution exists.
Let $(x,y,z)\in\Bbb{Z}^3$ be a solution to $(1)$. After changing signs if necessary we have $x,y,z\geq0$.
Observation 1: Without loss of generality $\gcd(x,y)=1$.
Proof. Let $d:=\gcd(x,y)$ so that $d^4$ divides $$x^4+4y^4=z^2,$$ and hence $d^2$ divides $z$. Setting $u:=\tfrac xd$, $v:=\tfrac yd$ and $w:=\tfrac{z}{d^2}$ we get another solution $(u,v,w)\in\Bbb{Z}^3$ to equation $(1)$, and we have $\gcd(u,v)=1$.$\qquad\square$
Observation 2: Without loss of generality $x$ and $z$ are odd, and $y$ is even.
Proof. Reducing mod $2$ shows that $x\equiv z\pmod{2}$. If both are odd then reducing mod $8$ shows that $y$ is even, and we are done. If both are even then setting $u:=\tfrac x2$ and $w=\tfrac z2$ shows that $$4u^4+y^4=w^2,$$ and so $(y,u,w)\in\Bbb{Z}^3$ is another solution. As before $y\equiv w\pmod{2}$, and because $\gcd(x,y)=1$ and $x$ is even, we find that $y$ and $w$ are odd, and then as before $u$ is even.$\qquad\square$
Theorem: There are no nontrivial solutions to the diophantine equation $$X^4-Y^4=Z^2.$$
Proof. This is theorem 114 on page 227 of Introduction to number theory by T. Nagell.
Observation 3: The solution $(x,y,z)\in\Bbb{Z}^3$ is trivial.
Proof. Because the triplet $(x,y,z)\in\Bbb{Z}^3$ satisfies equation $(1)$ we also have $$x^4=z^2-4y^4=(z+2y^2)(z-2y^2).$$ Let $d$ denote the $\gcd$ of the two factors on the right hand side. Then $d$ divides the difference $4y^2$ and the product $x^4$ of the two factors. Because $x$ is odd and $\gcd(x,y)=1$ it follows that $d=1$. This means both factors are fourth powers, say $$z+2y^2=u^4\qquad\text{ and }\qquad z-2y^2=v^4,$$ for some $u,v,\in\Bbb{Z}$. Then $(u,v,2y)\in\Bbb{Z}^3$ is a solution to the diophantine equation $$X^4-Y^4=Z^2.$$ Then $uvy=0$, and if any of these three equals $0$ it is not hard to see that $xyz=0$.$\qquad\square$
Conclusion: From an arbitrary solution $(x,y,z)\in\Bbb{Z}^3$ we have constructed a solution that turns out to be trivial. Tracing the construction back shows that the original solution is itself trivial. Hence there exist no nontrivial solutions to the diophantine equation $(1)$.