Show that the diophantine equation x^4+4y^4=z^2 has no solutions in nonzero integers
I think this might work.
Write $\displaystyle (x^2)^2+(2y^2)^2 = z^2$. Assume a non-zero solution exists. Let $\displaystyle z$ be the smallest of all positive solutions. Then it must mean $\displaystyle \gcd(x,y) = \gcd(x,z) = \gcd(y,z)=1$ by minimality of $\displaystyle z$. Using Pythagorean triples construct a smaller solution (that is, use Fermat's method of infinite descent) $\displaystyle z'<z$. And then this leads to a contradiction.