Let $P(x)=x^3+ax^2+b$ and $Q(x)=x^3+bx+a$, where $a$ and $b$ are nonzero real numbers.  Suppose that the roots of the equation $P(x)=0$ are the reciprocals of the roots of the equation $Q(x)=0$.  Prove that $a$ and $b$ are integers.  Find the greatest common divisor of $P(2013!+1)$ and $Q(2013!+1)$.
1. **Identify the roots and their reciprocals:**
   Let \( p, q, r \) be the roots of \( P(x) = 0 \). Since the roots of \( P(x) = 0 \) are the reciprocals of the roots of \( Q(x) = 0 \), we can denote the roots of \( Q(x) = 0 \) as \( \frac{1}{p}, \frac{1}{q}, \frac{1}{r} \).

2. **Express \( P(x) \) and \( Q(x) \) in terms of their roots:**
   By Vieta's formulas, for the polynomial \( P(x) = x^3 + ax^2 + b \), we have:
   \[
   p + q + r = -a, \quad pq + qr + rp = 0, \quad pqr = -b
   \]
   Similarly, for the polynomial \( Q(x) = x^3 + bx + a \), we have:
   \[
   \frac{1}{p} + \frac{1}{q} + \frac{1}{r} = 0, \quad \frac{1}{pq} + \frac{1}{qr} + \frac{1}{rp} = b, \quad \frac{1}{pqr} = -a
   \]

3. **Relate the coefficients using the roots:**
   From the roots of \( Q(x) \), we get:
   \[
   \frac{1}{p} + \frac{1}{q} + \frac{1}{r} = \frac{pq + qr + rp}{pqr} = 0 \implies pq + qr + rp = 0
   \]
   This is consistent with the condition for \( P(x) \).

   Also, we have:
   \[
   \frac{1}{pq} + \frac{1}{qr} + \frac{1}{rp} = \frac{p + q + r}{pqr} = \frac{-a}{-b} = \frac{a}{b}
   \]
   Since this must equal \( b \), we get:
   \[
   \frac{a}{b} = b \implies a = b^2
   \]

4. **Determine the values of \( a \) and \( b \):**
   Since \( a \) and \( b \) are nonzero real numbers, and \( a = b^2 \), \( b \) must be a nonzero real number. For \( a \) and \( b \) to be integers, \( b \) must be an integer, and hence \( a \) must be a perfect square. Therefore, \( a \) and \( b \) are integers.

5. **Find the greatest common divisor of \( P(2013! + 1) \) and \( Q(2013! + 1) \):**
   Given \( a = b^2 \), we can simplify the polynomials:
   \[
   P(x) = x^3 + b^2 x^2 + b, \quad Q(x) = x^3 + bx + b^2
   \]
   We need to find \( \gcd(P(2013! + 1), Q(2013! + 1)) \).

   Substitute \( x = 2013! + 1 \):
   \[
   P(2013! + 1) = (2013! + 1)^3 + b^2 (2013! + 1)^2 + b
   \]
   \[
   Q(2013! + 1) = (2013! + 1)^3 + b(2013! + 1) + b^2
   \]

   Since \( 2013! \) is divisible by all integers from 1 to 2013, \( 2013! + 1 \) is congruent to 1 modulo any integer \( k \) from 1 to 2013. Therefore, both \( P(2013! + 1) \) and \( Q(2013! + 1) \) are congruent to:
   \[
   1^3 + b^2 \cdot 1^2 + b = 1 + b^2 + b
   \]
   \[
   1^3 + b \cdot 1 + b^2 = 1 + b + b^2
   \]

   Hence, \( \gcd(P(2013! + 1), Q(2013! + 1)) = \gcd(1 + b + b^2, 1 + b + b^2) = 1 + b + b^2 \).

The final answer is \( \boxed{1 + b + b^2} \).