4. Solve the equation in integers $x^{2}+y^{2}=3 x y$.

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Answer: $x=y=0$.

Solution: If both numbers are not equal to 0, divide the numbers $x$ and $y$ by their greatest common divisor, resulting in coprime numbers $a$ and $b$. The right side of the equation is divisible by 3, so the left side must also be. The square of an integer can give a remainder of 0 or 1 when divided by 3, so $a^{2}$ and $b^{2}$ are divisible by 3, which means $a$ and $b$ are divisible by 3. This is a contradiction because $a$ and $b$ are coprime and have a common divisor 3, which is greater than 1.

Criteria: Only for the correct answer - 1 point. Noted that $x$ and $y$ must be divisible by $3-3$ points.