Problem 1. The squares of two numbers whose arithmetic mean is equal to 18, differ by 288. What are these numbers?
Solution. Let's denote the numbers by $x$ and $y$. From the condition of the problem, we have

$$
\frac{x+y}{2}=18 \text { and } x^{2}-y^{2}=288 \text {, i.e., } x+y=36 \text { and }(x-y)(x+y)=288 \text {. }
$$

If we substitute the first equation into the second, we get $36(x-y)=288$, i.e., $x-y=8$. Further, if we add the equations $x+y=36$ and $x-y=8$, we get $2 x=44$, i.e., $x=22$.

Finally, by substituting into $x+y=36$, we find $y=14$, which means the desired numbers are 22 and 14.