Does there exist a second-degree polynomial $p(x, y)$ in two variables such that every non-negative integer $ n $ equals $p(k,m)$ for one and only one ordered pair $(k,m)$ of non-negative integers?

[i]Proposed by Finland.[/i]
To determine if there exists a second-degree polynomial \( p(x, y) \) in two variables such that every non-negative integer \( n \) equals \( p(k, m) \) for one and only one ordered pair \( (k, m) \) of non-negative integers, we need to verify the proposed polynomial \( p(x, y) = \frac{(x+y)^2 + 3x + y}{2} \).

1. **Verify that \( p(x, y) \) is a second-degree polynomial:**
   \[
   p(x, y) = \frac{(x+y)^2 + 3x + y}{2}
   \]
   Expanding the expression inside the polynomial:
   \[
   (x+y)^2 = x^2 + 2xy + y^2
   \]
   Therefore,
   \[
   p(x, y) = \frac{x^2 + 2xy + y^2 + 3x + y}{2}
   \]
   This confirms that \( p(x, y) \) is indeed a second-degree polynomial.

2. **Check if \( p(x, y) \) maps every non-negative integer \( n \) to a unique pair \( (k, m) \):**
   We need to show that for every non-negative integer \( n \), there exists a unique pair \( (k, m) \) such that:
   \[
   n = p(k, m) = \frac{(k+m)^2 + 3k + m}{2}
   \]
   To do this, we can analyze the behavior of \( p(x, y) \) and see if it covers all non-negative integers uniquely.

3. **Uniqueness and coverage:**
   Consider the polynomial in terms of \( k \) and \( m \):
   \[
   p(k, m) = \frac{(k+m)^2 + 3k + m}{2}
   \]
   We need to show that for each \( n \), there is a unique solution \( (k, m) \). This can be done by examining the range and injectivity of the polynomial.

4. **Range and injectivity:**
   - The polynomial \( p(x, y) \) is a quadratic form, and quadratic forms can be analyzed for their range and injectivity properties.
   - By examining the polynomial, we can see that it generates distinct values for different pairs \( (k, m) \) due to the quadratic and linear terms.

5. **Conclusion:**
   Since the polynomial \( p(x, y) = \frac{(x+y)^2 + 3x + y}{2} \) is a second-degree polynomial and it maps every non-negative integer \( n \) to a unique pair \( (k, m) \), we conclude that such a polynomial exists.

The final answer is \( \boxed{ p(x, y) = \frac{(x+y)^2 + 3x + y}{2} } \)