Problem 1496

1. Assume for contradiction that the set $\{(x,y) \mid P(x,y) = 0\}$ is bounded. This means there exists a positive number $M$ such that the distance of any point $(x,y)$ in this set from the origin is less than $M$.

2. Identify the points $(x_1, y_1)$ and $(x_2, y_2)$ such that $Q(x_1, y_1) > 0$ and $Q(x_2, y_2) < 0$. Since $Q(x,y)$ is the sum of the monomials of the highest degree in $P(x,y)$, it is a homogeneous polynomial of degree $d$ (the degree of $P(x,y)$).

3. Consider the polynomials $P(tx_1, ty_1)$ and $P(tx_2, ty_2)$ for a scalar $t$. Since $Q(x,y)$ is homogeneous of degree $d$, we have:
$$Q(tx_1, ty_1) = t^d Q(x_1, y_1) \quad \text{and} \quad Q(tx_2, ty_2) = t^d Q(x_2, y_2)$$
Given $Q(x_1, y_1) > 0$ and $Q(x_2, y_2) < 0$, it follows that:
$$Q(tx_1, ty_1) > 0 \quad \text{and} \quad Q(tx_2, ty_2) < 0 \quad \text{for any } t > 0$$

4. **Analyze the behavior of $P(tx_1, ty_1)$ and $P(tx_2, ty_2)$** as $t \to \infty$. Since $Q(x,y)$ represents the leading term of $P(x,y)$, for sufficiently large $t$, the sign of $P(tx_1, ty_1)$ will be dominated by $Q(tx_1, ty_1)$, and similarly for $P(tx_2, ty_2)$. Therefore:
$$P(tx_1, ty_1) > 0 \quad \text{and} \quad P(tx_2, ty_2) < 0 \quad \text{for sufficiently large } t$$

5. **Choose a sufficiently large $t = s$** such that the points $(sx_1, sy_1)$ and $(sx_2, sy_2)$ lie outside the circle of radius $M$ centered at the origin. This is possible because as $t \to \infty$, the points $(tx_1, ty_1)$ and $(tx_2, ty_2)$ move further away from the origin.

6. **Construct a continuous path $C$** from $(sx_1, sy_1)$ to $(sx_2, sy_2)$ that lies outside the circle of radius $M$. Since $P(x,y)$ is continuous, the restriction of $P$ to the path $C$ is also continuous.

7. Apply the Intermediate Value Theorem to the continuous function $P$ on the path $C$. Since $P(sx_1, sy_1) > 0$ and $P(sx_2, sy_2) < 0$, there must exist a point $(x, y) \in C$ such that $P(x, y) = 0$.

8. Contradiction: The point $(x, y) \in C$ where $P(x, y) = 0$ lies outside the circle of radius $M$, contradicting the assumption that the set $\{(x,y) \mid P(x,y) = 0\}$ is bounded.

Therefore, the set $\{(x,y) \mid P(x,y) = 0\}$ is not bounded.

$\blacksquare$