3. Solve the equation in integers $x^{4}-2 y^{4}-4 z^{4}-8 t^{4}=0$.

## Answer: $x=y=z=t=0$
Note that $x$ is even. Let $x=2x_{1}$, then we get

![](https://cdn.mathpix.com/cropped/2024_05_06_1b3c559ebe88f2ab36bbg-1.jpg?height=52&width=955&top_left_y=951&top_left_x=1581)
$4x_{1}^{4}-8y_{1}^{4}-z^{4}-2t^{4}=0$. Therefore, $z=2z_{1}$ and $2x_{1}^{4}-4y_{1}^{4}-8z_{1}^{4}-t^{4}=0$. Similarly, $t=2t_{1}$ and $x_{1}^{4}-2y_{1}^{4}-4z_{1}^{4}-8t_{1}^{4}=0$ we obtain the original equation, but for variables $x_{1}, y_{1}, z_{1}, t_{1}$, so we can repeat the same reasoning and so on. We get that each variable is divisible by any power of 2. Therefore, $x=y=z=t=0$.

## Recommendations for checking

1) Noted that $x$ is even - 1 point.
2) Using correct reasoning, the equation

$x_{1}^{4}-2y_{1}^{4}-4z_{1}^{4}-8t_{1}^{4}=0$, where $x=2x_{1}, y=2y_{1}, z=2z_{1}, t=2t_{1}$, but no corresponding conclusion - 4 points.

3) Only the answer is provided without proof that there are no other solutions - 0 points.