Problem 1428

1. We start with the given functional equation for the polynomial $P(x, y)$:
$$P(x - 1, y - 2x + 1) = P(x, y).$$

2. We will use induction to show that $P(x, y)$ can be expressed in terms of $P(0, y - x^2)$.

3. Consider the polynomial evaluated at $x = n$ for $n \in \mathbb{N}$:
$$P(n, y) = P(n - 1, y - 2n + 1).$$

4. Applying the functional equation repeatedly, we get:
$$P(n, y) = P(n - 1, y - 2n + 1) = P(n - 2, y - 2(n - 1) + 1 - 2n + 1).$$

5. Continuing this process, we have:
$$P(n, y) = P(n - 2, y - (2n - 1) - (2(n - 1) - 1)).$$

6. Generalizing this, we get:
$$P(n, y) = P(0, y - (2n - 1) - (2(n - 1) - 1) - \cdots - 1).$$

7. The sum of the first $n$ odd numbers is $n^2$, so:
$$P(n, y) = P(0, y - n^2).$$

8. Define $Q_y(x) = P(x, y) - P(0, y - x^2)$. Notice that $Q_y(x)$ is a polynomial in $x$.

9. From the previous steps, we know that $Q_y(n) = 0$ for all $n \in \mathbb{N}$.

10. Since $Q_y(x)$ is a polynomial and has infinitely many roots (all natural numbers), it must be the zero polynomial. Therefore:
$$Q_y(x) = 0 \quad \forall x \in \mathbb{R}.$$

11. This implies:
$$P(x, y) = P(0, y - x^2) \quad \forall x, y \in \mathbb{R}.$$

12. Let $\Phi(t) = P(0, t)$. Then we can write:
$$P(x, y) = \Phi(y - x^2).$$

Thus, we have shown that there exists a polynomial $\Phi(x)$ such that $P(x, y) = \Phi(y - x^2)$.

$\blacksquare$