Determine all pairs of non-constant polynomials $p(x)$ and $q(x)$, each with leading coefficient $1$, degree $n$, and $n$ roots which are non-negative integers, that satisfy $p(x)-q(x)=1$.
Given the problem, we need to determine all pairs of non-constant polynomials \( p(x) \) and \( q(x) \), each with leading coefficient 1, degree \( n \), and \( n \) roots which are non-negative integers, that satisfy \( p(x) - q(x) = 1 \).

1. **Express \( p(x) \) in terms of \( q(x) \):**
   \[
   p(x) = q(x) + 1
   \]

2. **Assume the forms of \( p(x) \) and \( q(x) \):**
   \[
   p(x) = (x - \alpha_1)(x - \alpha_2) \cdots (x - \alpha_n)
   \]
   \[
   q(x) = (x - \beta_1)(x - \beta_2) \cdots (x - \beta_n)
   \]
   where \( \alpha_i \) and \( \beta_i \) are non-negative integers.

3. **Case 1: \( \alpha_n \ge \beta_i \) for all \( 1 \le i \le n \):**
   - Consider \( q(\alpha_n) \):
     \[
     q(\alpha_n) = \prod_{i=1}^n (\alpha_n - \beta_i)
     \]
   - Since \( p(x) = q(x) + 1 \), we have:
     \[
     p(\alpha_n) = q(\alpha_n) + 1 = 0 + 1 = 1
     \]
     \[
     \prod_{i=1}^n (\alpha_n - \beta_i) = -1
     \]
   - This implies that \( \beta_1 = \beta_2 = \cdots = \beta_n = \beta \) (a constant integer), and:
     \[
     q(x) = (x - \beta)^n
     \]
     \[
     p(x) = (x - \beta)^n + 1
     \]
   - For \( p(x) \) to have integer roots, \( n \) must be odd. 

4. **Check the constraints using Cauchy-Schwarz inequality:**
   - Sum of products of roots:
     \[
     \sum_{i \neq j} \alpha_i \alpha_j = \frac{n(n-1)\beta^2}{2}
     \]
   - Sum of squares of roots:
     \[
     \sum \alpha_i^2 = n \beta^2
     \]
   - Applying Cauchy-Schwarz inequality:
     \[
     \left( \sum_{i \neq j} \alpha_i \alpha_j \right)^2 \le \left( \sum \alpha_i^2 \right)^2
     \]
     \[
     \left( \frac{n(n-1)\beta^2}{2} \right)^2 \le (n \beta^2)^2
     \]
     \[
     \frac{(n(n-1))^2 \beta^4}{4} \le n^2 \beta^4
     \]
     \[
     (n-1)^2 \le 4
     \]
     \[
     n \le 3
     \]
   - Since \( n \) must be odd, \( n = 1 \) is the only solution that works. For \( n = 3 \), \( (x - \beta)^3 + 1 \) does not have all integer roots.

5. **Case 2: \( \beta_n \ge \alpha_i \) for all \( 1 \le i \le n \):**
   - Consider \( p(\beta_n) \):
     \[
     p(\beta_n) = \prod_{i=1}^n (\beta_n - \alpha_i)
     \]
   - Since \( p(x) = q(x) + 1 \), we have:
     \[
     p(\beta_n) = q(\beta_n) + 1 = 0 + 1 = 1
     \]
     \[
     \prod_{i=1}^n (\beta_n - \alpha_i) = 1
     \]
   - This implies that \( \alpha_1 = \alpha_2 = \cdots = \alpha_n = \alpha \) (a constant integer), and:
     \[
     p(x) = (x - \alpha)^n
     \]
     \[
     q(x) = (x - \alpha)^n - 1
     \]
   - For \( q(x) \) to have integer roots, \( n \) must be even.

6. **Check the constraints using Cauchy-Schwarz inequality:**
   - Sum of products of roots:
     \[
     \sum_{i \neq j} \beta_i \beta_j = \frac{n(n-1)\alpha^2}{2}
     \]
   - Sum of squares of roots:
     \[
     \sum \beta_i^2 = n \alpha^2
     \]
   - Applying Cauchy-Schwarz inequality:
     \[
     n \le 3
     \]
   - Since \( n \) must be even, \( n = 2 \) is the only solution that works.

Conclusion:
Combining all solutions, we get:
\[
\boxed{(p(x) = x - \beta + 1, q(x) = x - \beta), (p(x) = (x - \alpha)^2, q(x) = (x - \alpha)^2 - 1)}
\]
for some \( \alpha, \beta \in \mathbb{Z}^* \).