10.5. What is the smallest value that the GCD $(x, y)$ can take, where $x$ and $y$ are natural numbers, if the LCM $(x, y)=(x-y)^{2}$?
Solution:

When $x=4$ and $y=2$, LCM $(x, y)=(x-y)^{2}=4$, and GCD $(x, y)=2$. Thus, 2 is achievable.

Assume that GCD $(x, y)=1$ (the numbers are coprime). Then LCM $(x, y)=x y$. From the condition, we get that $x y=x^{2}-2 x y+y^{2}$. We obtain $x^{2}-3 x y+y^{2}=0$. Solving the obtained quadratic equation with respect to $x$, we get the discriminant $5 y^{2}$. The square root of the discriminant $\sqrt{5} \cdot y$ is an irrational number, and for $x$ to be a natural number, it must be rational. This is a contradiction. Therefore, GCD $(x, y) \neq 1$.

Answer: 2.

| Criteria | Points |
| :--- | :---: |
| Complete solution of the problem. | 7 |
| Proved that GCD $(x, y) \neq 1$. Correct answer, but no example of numbers for GCD $(x, y)=2$ provided. | 5 |
| Correct answer with an example. | 2 |
| Only the correct answer, without evaluation and example. | 0 |