Find all solutions to $aabb=n^4-6n^3$, where $a$ and $b$ are non-zero digits, and $n$ is an integer.  ($a$ and $b$ are not necessarily distinct.)
To find all solutions to the equation \(aabb = n^4 - 6n^3\), where \(a\) and \(b\) are non-zero digits and \(n\) is an integer, we can proceed as follows:

1. **Rewrite \(aabb\) in a more convenient form:**
   \[
   aabb = 1100a + 11b
   \]
   This is because \(aabb\) can be interpreted as \(1000a + 100a + 10b + b = 1100a + 11b\).

2. **Set up the equation:**
   \[
   1100a + 11b = n^4 - 6n^3
   \]

3. **Factor out the common factor on the left-hand side:**
   \[
   11(100a + b) = n^4 - 6n^3
   \]
   This implies that \(11\) must divide \(n^4 - 6n^3\).

4. **Check divisibility by 11:**
   \[
   n^4 - 6n^3 \equiv 0 \pmod{11}
   \]
   This means \(n^3(n - 6) \equiv 0 \pmod{11}\). Therefore, \(n\) must be a multiple of 11 or \(n \equiv 6 \pmod{11}\).

5. **Consider the two cases:**
   - **Case 1: \(n = 11k\) for some integer \(k\):**
     \[
     n = 11
     \]
     Plugging \(n = 11\) into the equation:
     \[
     n^4 - 6n^3 = 11^4 - 6 \cdot 11^3 = 14641 - 7986 = 6655
     \]
     We need to check if \(6655\) can be written as \(aabb\):
     \[
     6655 = 1100a + 11b
     \]
     Solving for \(a\) and \(b\):
     \[
     6655 = 1100 \cdot 6 + 11 \cdot 5
     \]
     Thus, \(a = 6\) and \(b = 5\), so \(aabb = 6655\).

   - **Case 2: \(n = 11k + 6\) for some integer \(k\):**
     \[
     n = 17
     \]
     Plugging \(n = 17\) into the equation:
     \[
     n^4 - 6n^3 = 17^4 - 6 \cdot 17^3 = 83521 - 6 \cdot 4913 = 83521 - 29478 = 54043
     \]
     We need to check if \(54043\) can be written as \(aabb\):
     \[
     54043 \neq 1100a + 11b
     \]
     Since \(54043\) is not in the form \(aabb\), this case does not provide a valid solution.

6. **Conclusion:**
   The only valid solution is when \(n = 11\), which gives \(aabb = 6655\).

The final answer is \(\boxed{6655}\).