The polynomial $A(x) = x^2 + ax + b$ with integer coefficients has the following property: 
for each prime $p$ there is an integer $k$ such that $A(k)$ and $A(k + 1)$ are both divisible by $p$. 
Proof that there is an integer $m$ such that $A(m) = A(m + 1) = 0$.
1. Given the polynomial \( A(x) = x^2 + ax + b \) with integer coefficients, we know that for each prime \( p \), there exists an integer \( k \) such that both \( A(k) \) and \( A(k+1) \) are divisible by \( p \). This means:
   \[
   p \mid A(k) \quad \text{and} \quad p \mid A(k+1)
   \]

2. We can write:
   \[
   A(k) = k^2 + ak + b
   \]
   \[
   A(k+1) = (k+1)^2 + a(k+1) + b = k^2 + 2k + 1 + ak + a + b
   \]

3. Since \( p \mid A(k) \) and \( p \mid A(k+1) \), we have:
   \[
   p \mid A(k+1) - A(k)
   \]

4. Calculating \( A(k+1) - A(k) \):
   \[
   A(k+1) - A(k) = (k^2 + 2k + 1 + ak + a + b) - (k^2 + ak + b) = 2k + 1 + a
   \]

5. Therefore:
   \[
   p \mid 2k + 1 + a
   \]

6. This implies:
   \[
   2k \equiv -1 - a \pmod{p}
   \]

7. Squaring both sides, we get:
   \[
   4k^2 \equiv (-1 - a)^2 \pmod{p}
   \]

8. Since \( p \mid A(k) \), we have:
   \[
   p \mid k^2 + ak + b
   \]

9. Multiplying by 4:
   \[
   4k^2 + 4ak + 4b \equiv 0 \pmod{p}
   \]

10. Substituting \( 4k^2 \equiv (-1 - a)^2 \):
    \[
    (-1 - a)^2 + 4ak + 4b \equiv 0 \pmod{p}
    \]

11. Simplifying:
    \[
    a^2 + 2a + 1 + 4b \equiv 0 \pmod{p}
    \]

12. This must hold for all primes \( p \), so:
    \[
    a^2 + 2a + 1 + 4b = 0
    \]

13. Solving for \( b \):
    \[
    a^2 + 2a + 1 = -4b
    \]
    \[
    a^2 + 2a + 1 = 4b
    \]

14. Recognizing the left-hand side as a perfect square:
    \[
    (a + 1)^2 = 4b
    \]

15. Let \( a = 2c + 1 \) for some integer \( c \):
    \[
    (2c + 1 + 1)^2 = 4b
    \]
    \[
    (2c + 2)^2 = 4b
    \]
    \[
    4(c + 1)^2 = 4b
    \]
    \[
    b = (c + 1)^2
    \]

16. Substituting \( a = 2c + 1 \) and \( b = c^2 + c \) into \( A(x) \):
    \[
    A(x) = x^2 + (2c + 1)x + (c^2 + c)
    \]
    \[
    A(x) = (x + c)(x + c + 1)
    \]

17. Therefore, there exists an integer \( m = -c \) such that:
    \[
    A(m) = A(m + 1) = 0
    \]

The final answer is \( \boxed{(x + c)(x + c + 1)} \)