45. Given an arithmetic progression consisting of four integers. It is known that the largest of these numbers is equal to the sum of the squares of the other three numbers. Find these numbers.
45. Let $x$ be the largest term of the progression, then the second is $x-d$, the third is $x-2d$, and the fourth is $x-3d$. The condition of the problem can be written as:

$$
x=(x-d)^{2}+(x-2d)^{2}+(x-3d)^{2}
$$

From this, $d=\frac{6x \pm \sqrt{2x(7-3x)}}{14}, x>0, 7-3x \geqslant 0, x \leqslant 2 \frac{1}{3}$.

Thus, $0<x \leqslant 2 \frac{1}{3}$. But $x$ is an integer, so $x=1, x=2$. Only $x=2, d=1$ satisfies the condition of the problem.

The required progression is $2,1,0,-1$.