19-th Macedonian Mathematical Olympiad

It is clear that $(0,0,0,0)$ is a solution of the equation. We will show that there exists no nonzero solution of the equation. Let us suppose the contrary, i.e. let $(x_0,y_0,z_0,t_0)$ be a solution of the equation with at least one nonzero coordinate $x_0^4+2y_0^4+4z_0^4+8t_0^4=16x_0{y}_0{z}_0{t}_0$. It is clear that $x_0$ is divisible by $2$. Therefore we write it in the form $x_0=2x_1$ for some $x_1\in \mathbb{Z}$, we substitute in the first equation and we divide by $2$, after which we get the equation $y_0^4+2z_0^4+4t_0^4+8x_1^4=16x_1{y}_0{z}_0{t}_0$. Now the number $y_0$ is divisible by $2$ so $y_0=2y_1$ for some $y_1\in \mathbb{Z}$. We again substitute in the last equation and we divide by $2$, after which we get $z_0^4+2t_0^4+4x_1^4+8y_1^4=16x_1{y}_1{z}_0{t}_0$. Analogously, the number $z_0$ is divisible by $2$, so $z_0=2z_1$ for $z_1\in \mathbb{Z}$. We substitute in the last equation and divide by $2$, after which we get the equation $t_0^4+2x_1^4+4y_1^4+8z_1^4=16x_1{y}_1{z}_1{t}_0$. The number $t_0$ is divisible by $2$ i.e. $t_0=2t_1$ for $t_1\in \mathbb{Z}$. After substitution and division by $2$ we get the equation $x_1^4+2y_1^4+4z_1^4+8t_1^4=16x_1{y}_1{z}_1{t}_1$. We get that the quadruple $(x_1,y_1,z_1,t_1)$ is also a solution of the first equation. If we continue this procedure, we get quadruples of integer solutions of the first equation $(x_2,y_2,z_2,t_2)$, $(x_3,y_3,z_3,t_3)$, with the solution at the $n$-th step being $(x_n,y_n,z_n,t_n)=\left(\frac{x_0}{2^n},\frac{y_0}{2^n},\frac{z_0}{2^n},\frac{t_0}{2^n}\right)$, which contradicts the fact the solutions are integer quadruples. Therefore the only solution is $(0,0,0,0)$.