For which integers $n = 1, \ldots  , 6$ does the equation $$a^n + b^n = c^n + n$$ have a solution in integers?

1. **For \( n = 1 \):**
   \[
   a^1 + b^1 = c^1 + 1 \implies a + b = c + 1
   \]
   We can take \( (a, b, c) = (1, 1, 1) \):
   \[
   1 + 1 = 1 + 1 \implies 2 = 2
   \]
   Thus, the equation has a solution for \( n = 1 \).

2. **For \( n = 2 \):**
   \[
   a^2 + b^2 = c^2 + 2
   \]
   We can take \( (a, b, c) = (1, 1, 0) \):
   \[
   1^2 + 1^2 = 0^2 + 2 \implies 1 + 1 = 0 + 2 \implies 2 = 2
   \]
   Thus, the equation has a solution for \( n = 2 \).

3. **For \( n = 3 \):**
   \[
   a^3 + b^3 = c^3 + 3
   \]
   We can take \( (a, b, c) = (1, 1, -1) \):
   \[
   1^3 + 1^3 = (-1)^3 + 3 \implies 1 + 1 = -1 + 3 \implies 2 = 2
   \]
   Thus, the equation has a solution for \( n = 3 \).

4. **For \( n = 4 \):**
   \[
   a^4 + b^4 = c^4 + 4
   \]
   Consider the equation modulo 4:
   \[
   a^4 + b^4 \equiv c^4 + 4 \pmod{4}
   \]
   Since \( x^4 \equiv 0 \) or \( 1 \pmod{4} \), we have:
   \[
   a^4 + b^4 \equiv 0, 1, 2 \pmod{4}
   \]
   and
   \[
   c^4 + 4 \equiv 0, 1, 2, 3 \pmod{4}
   \]
   For \( a^4 + b^4 \equiv c^4 + 4 \pmod{4} \), we need:
   \[
   a^4 + b^4 \equiv c^4 \pmod{4}
   \]
   This implies \( a \equiv b \equiv c \equiv 0 \pmod{2} \). Then:
   \[
   a^4 + b^4 \equiv 0 \pmod{8} \quad \text{and} \quad c^4 + 4 \equiv 4 \pmod{8}
   \]
   This is a contradiction. Thus, the equation has no solution for \( n = 4 \).

5. **For \( n = 5 \):**
   \[
   a^5 + b^5 = c^5 + 5
   \]
   Consider the equation modulo 11:
   \[
   a^5 + b^5 \equiv c^5 + 5 \pmod{11}
   \]
   Since \( x^5 \equiv -1, 0, 1 \pmod{11} \), we have:
   \[
   a^5 + b^5 \equiv -2, -1, 0, 1, 2 \pmod{11}
   \]
   and
   \[
   c^5 + 5 \equiv -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6 \pmod{11}
   \]
   This is a contradiction. Thus, the equation has no solution for \( n = 5 \).

6. **For \( n = 6 \):**
   \[
   a^6 + b^6 = c^6 + 6
   \]
   Consider the equation modulo 8:
   \[
   a^6 + b^6 \equiv c^6 + 6 \pmod{8}
   \]
   Since \( x^6 \equiv 0, 1 \pmod{8} \), we have:
   \[
   a^6 + b^6 \equiv 0, 1, 2 \pmod{8}
   \]
   and
   \[
   c^6 + 6 \equiv 6, 7 \pmod{8}
   \]
   This is a contradiction. Thus, the equation has no solution for \( n = 6 \).

The final answer is \( \boxed{ n = 1, 2, 3 } \).