Determine all pairs of polynomials $(P, Q)$ with real coefficients satisfying
$$P(x + Q(y)) = Q(x + P(y))$$
for all real numbers $x$ and $y$.
1. Given the functional equation:
   \[
   P(x + Q(y)) = Q(x + P(y))
   \]
   for all real numbers \(x\) and \(y\).

2. Let's analyze the equation by considering the degrees of the polynomials \(P\) and \(Q\). Suppose \(P\) and \(Q\) are polynomials of degree \(n\) and \(m\) respectively.

3. By comparing the degrees on both sides of the equation, we see that the degree of \(P(x + Q(y))\) must equal the degree of \(Q(x + P(y))\). This implies:
   \[
   \deg(P) = \deg(Q)
   \]
   Let \(d\) be the common degree of \(P\) and \(Q\).

4. Consider the case when \(d = 0\). This means both \(P\) and \(Q\) are constant polynomials. Let \(P(x) = c_1\) and \(Q(x) = c_2\) for some constants \(c_1\) and \(c_2\). Substituting these into the original equation, we get:
   \[
   c_1 = c_2
   \]
   Hence, \(P\) and \(Q\) must be the same constant polynomial.

5. Now consider the case when \(d > 0\). Suppose \(P(x) = a_d x^d + \cdots + a_0\) and \(Q(x) = b_d x^d + \cdots + b_0\). Since the degrees are equal, the leading coefficients must also be equal for the equation to hold for all \(x\) and \(y\). Therefore, \(a_d = b_d\).

6. Next, we consider the case where \(P(x) - Q(x)\) is a non-constant polynomial. Let \(R(x) = P(x) - Q(x)\). Then \(R(x)\) is a polynomial of degree \(d\). Substituting \(P(x) = Q(x) + R(x)\) into the original equation, we get:
   \[
   Q(x + Q(y) + R(y)) = Q(x + Q(y) + R(y))
   \]
   This equation holds trivially, but it does not provide any new information. Therefore, we need to consider the specific forms of \(P\) and \(Q\).

7. If \(P(x) - Q(x)\) is a constant polynomial, say \(P(x) - Q(x) = t\) for some constant \(t\), then:
   \[
   P(x) = Q(x) + t
   \]
   Substituting this into the original equation, we get:
   \[
   Q(x + Q(y) + t) = Q(x + Q(y) + t)
   \]
   This equation holds for all \(x\) and \(y\), implying that \(P(x) = Q(x) + t\) is a valid solution.

8. To find specific forms of \(P\) and \(Q\), consider the simplest non-trivial case where \(P(x) = x + a\) and \(Q(x) = x + b\) for some constants \(a\) and \(b\). Substituting these into the original equation, we get:
   \[
   (x + a) + (y + b) = (x + b) + (y + a)
   \]
   Simplifying, we see that this equation holds for all \(x\) and \(y\).

Conclusion:
The pairs of polynomials \((P, Q)\) that satisfy the given functional equation are:
\[
\boxed{P(x) = Q(x) \text{ or } P(x) = x + a, Q(x) = x + b \text{ for some constants } a \text{ and } b}
\]