Find all non-negative integers $x, y, z$ such that $x^{3}+4 y^{3}=16 z^{3}+4 x y z$.

## 2 Solutions
Let $x, y, z$ be a solution. We notice that $4 y^{3}, 16 z^{3}$, and $4 x y z$ are even, so $x^{3}$ must be even, and thus $x$ must be even. We set $x^{\prime}=x / 2$. The equation becomes $8 x^{13}+4 y^{3}=16 z^{2}+8 x^{\prime} y z$, or equivalently $2 x^{\prime 3}+y^{3}=4 z^{2}+2 x^{\prime} y z$. We notice that $y$ is even: we write $y=2 y^{\prime}$, and the equation becomes $x^{\prime 3}+4 y^{\prime 3}=2 z^{2}+2 x^{\prime} y^{\prime} z$. We notice that $x^{\prime}$ is even: we write $x^{\prime}=2 x^{\prime \prime}$, and the equation becomes $4 x^{\prime \prime 3}+2 y^{\prime 3}=z^{2}+2 x^{\prime \prime} y^{\prime} z$. We notice that $z$ is even: we write $z=2 z^{\prime}$, and the equation becomes $2 x^{\prime \prime 3}+y^{\prime 3}=4 z^{\prime 2}+2 x^{\prime \prime} y^{\prime} z^{\prime}$. We notice that this fifth equation is identical to the second! We have thus proven that, if $\left(x^{\prime}, y, z\right)$ is a solution to the equation $2 x^{\prime 3}+y^{3}=4 z^{2}+2 x^{\prime} y z$, which we will call (E), then $\left(x^{\prime} / 2, y / 2, z / 2\right)$ is also a solution. By the principle of infinite descent, the only solution to equation (E) is $(0,0,0)$. Since we had that, if $(x, y, z)$ is a solution to the original equation, $(x / 2, y, z)$ is a solution to (E). Thus, $(x / 2, y, z)=(0,0,0)$, and therefore $x=y=z=0$ as required.

## 2 Group B

## 1 Statements