1. In the set of integers, solve the equation

$$
x^{4}+2 y^{4}+4 z^{4}+8 t^{4}=16 x y z t
$$
Solution. Let us assume that $(x, y, z, t)$ is a solution to the given equation. Clearly, $2 \mid x$, i.e., $x=2a$ and if we substitute in (1), and then divide by 2, we get $y^{4}+2z^{4}+4t^{4}+8a^{4}=16ayzt$. But this means that $2 \mid y$, i.e., $y=2b$ and if we substitute in the previous equation, and then divide by 2, we get $z^{4}+2t^{4}+4a^{4}+8b^{4}=16abzt$. Now we get that $2 \mid z$, i.e., $z=2c$, and if we substitute in the previous equation, and then divide by 2, we get $t^{4}+2a^{4}+4b^{4}+8c^{4}=16abct$. But this means that $2 \mid t$, i.e., $t=2d$ and if we substitute in the previous equation, and then divide by 2, we get

$$
a^{4}+2b^{4}+4c^{4}+8d^{4}=16abcd
$$

i.e., the quadruple $(a, b, c, d)=\left(\frac{x}{2}, \frac{y}{2}, \frac{z}{2}, \frac{t}{2}\right)$ is a solution to (1). Continuing this process, we get that for every $n \in \mathbb{N}$, the ordered quadruple $\left(\frac{x}{2^{n}}, \frac{y}{2^{n}}, \frac{z}{2^{n}}, \frac{t}{2^{n}}\right)$ is an integer solution to (1). The latter means that each of the numbers $x, y, z, t$ is divisible by 2 an arbitrary number of times, which is possible if and only if $x=y=z=t=0$.