Problem 1. Let $n$ be a natural number. Find the integers $x, y, z$ with the property

$$
x^{2}+y^{2}+z^{2}=2^{n}(x+y+z)
$$
Solution. If $n=0$, using the obvious inequalities $x^{2} \geq x$ and their analogues,

![](https://cdn.mathpix.com/cropped/2024_06_07_8dca213771210a27e81cg-2.jpg?height=46&width=1211&top_left_y=671&top_left_x=457)

If $n \geq 1$, then 2 divides $x^{2}+y^{2}+z^{2}$, so either all three numbers are even, or one is even and two are odd. In the latter case, taking, for example, $x=2 x_{1}+1, y=2 y_{1}+1, z=2 z_{1}$, we obtain

$$
4\left(x_{1}^{2}+x_{1}+y_{1}^{2}+y_{1}+z_{1}^{2}\right)+2=4\left(x_{1}+y_{1}+z_{1}+1\right)
$$

contradiction $\qquad$
The remaining case is when $x, y, z$ are even. For $x=2 x_{1}, y=2 y_{1}, z=2 z_{1}$, we obtain

$$
x_{1}^{2}+y_{1}^{2}+z_{1}^{2}=2^{n-1}\left(x_{1}+y_{1}+z_{1}\right)
$$

![](https://cdn.mathpix.com/cropped/2024_06_07_8dca213771210a27e81cg-2.jpg?height=46&width=1211&top_left_y=1126&top_left_x=457)

For $n>1$, repeating the argument, we deduce that if $x=2^{n} x_{n}, y=$ $2^{n} y_{n}, z=2^{n} z_{n}$, then $x_{n}, y_{n}, z_{n} \in \mathbb{Z}$ and

$$
x_{n}^{2}+y_{n}^{2}+z_{n}^{2}=x_{n}+y_{n}+z_{n}
$$

from which $x_{n}, y_{n}, z_{n} \in\{0,1\}$, so $x, y, z \in\left\{0,2^{n}\right\}$ 2 points