Mediterranean Mathematical Competition

This is an easy problem with many solutions. We rewrite
$$\begin{aligned} 4x^4 + y^4 - z^2 + 4xyz &= (4x^4 + y^4 + 4x^2y^2) - (4x^2y^2 + z^2 - 4xyz) \\ &= (2x^2 + y^2)^2 - (2xy - z)^2 \\ &= (2x^2 + y^2 - 2xy + z)(2x^2 + y^2 + 2xy - z) \end{aligned}$$
The two factors $A = 2x^2 + y^2 - 2xy + z$ and $B = 2x^2 + y^2 + 2xy - z$ add up to $A + B = 4x^2 + 2y^2$. If we choose the values of $x$ and $y$ so that $A = 4x^2 = 4 \cdot 5^{2n+2}$ and $B = 2y^2 = 2 \cdot 2^{2n}$, then the product will become $AB = 2 \cdot 10^{2n+2}$ and the sum of the digits will equal 2.

Summarizing, we pick an integer $n \ge 1$ and set $x = 5^{n+1}$ and $y = 2^n$. The desired equation $A = 4x^2$ is equivalent to $2x^2 + y^2 - 2xy + z = 4x^2$, and hence
$$z = 2x^2 + 2xy - y^2 = 2 \cdot 5^{2n+2} + 10^{n+1} - 4^n.$$
Note that $z$ indeed is a positive integer. As the described choice of $x, y, z$ then yields
$$4x^4 + y^4 - z^2 + 4xyz = 2 \cdot 10^{2n+2},$$
the proof is complete.