1. **Define the sequence $b_{ij}$:**
We need to define a sequence $b_{ij}$ such that:
- $b_{ij} = b_{ji}$
- $\{i,j\} \neq \{m,n\} \implies \gcd(b_{ij}, b_{mn}) = 1$

2. **Construct the polynomial $P(x)$:**
Define $P(a_i) = \prod_{j=1}^{n+1} b_{ij}$. This ensures that $P(a_i)$ and $P(a_j)$ share common factors for $i \neq j$, satisfying the first condition.

3. Lagrange Interpolation:
By Lagrange interpolation, we can find a polynomial $P(x) \in \mathbb{Q}[x]$ of degree at most $n$ that satisfies $P(a_i) = \prod_{j=1}^{n+1} b_{ij}$ for $i = 1, 2, \ldots, n+1$.

4. **Ensure $P(x) \in \mathbb{Z}[x]$:**
We need to show that $P(x)$ can be chosen to have integer coefficients. By Lagrange interpolation, we have:
$$P(x) = \sum_{i=1}^{n+1} P(a_i) \prod_{\substack{1 \leq j \leq n+1 \\ j \neq i}} \left( \frac{x - a_j}{a_i - a_j} \right)$$
Define $A = \prod_{i=1}^{n+1} \left( \prod_{\substack{1 \leq j \leq n+1 \\ j \neq i}} (a_i - a_j) \right)$.

5. **Choose $b_{ij}$ such that $b_{ij} \equiv 1 \pmod{A}$:**
We need $\frac{n(n+1)}{2}$ relatively coprime integers congruent to 1 modulo $A$. This can be achieved by setting:
$$b_1 = 1, \quad b_i = A \prod_{j=1}^{i-1} b_j + 1 \quad \forall 2 \leq i \leq \frac{n(n+1)}{2}$$

6. **Verify $P(a_i) \equiv 1 \pmod{A}$:**
Set $k_i = \frac{P(a_i) - 1}{A}$. We need to show that $P(x) \in \mathbb{Z}[x]$. Using Lagrange interpolation on the constant polynomial 1, we have:
$$1 = \sum_{i=1}^{n+1} 1 \cdot \prod_{\substack{1 \leq j \leq n+1 \\ j \neq i}} \left( \frac{x - a_j}{a_i - a_j} \right)$$
This implies:
$$1 = \sum_{i=1}^{n+1} (P(a_i) - Ak_i) \frac{\prod_{\substack{1 \leq j \leq n+1 \\ j \neq i}} (x - a_j)}{\prod_{\substack{1 \leq j \leq n+1 \\ j \neq i}} (a_i - a_j)}$$
Simplifying, we get:
$$1 = P(x) - \sum_{i=1}^{n+1} (Ak_i) \frac{\prod_{\substack{1 \leq j \leq n+1 \\ j \neq i}} (x - a_j)}{\prod_{\substack{1 \leq j \leq n+1 \\ j \neq i}} (a_i - a_j)}$$
Since $\prod_{\substack{1 \leq j \leq n+1 \\ j \neq i}} (a_i - a_j) \mid A$, the term:
$$\sum_{i=1}^{n+1} (Ak_i) \frac{\prod_{\substack{1 \leq j \leq n+1 \\ j \neq i}} (x - a_j)}{\prod_{\substack{1 \leq j \leq n+1 \\ j \neq i}} (a_i - a_j)}$$
has integer coefficients, ensuring that $P(x) \in \mathbb{Z}[x]$.

$\blacksquare$