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).

Answer

To solve the equation \( aabb = n^4 - 6n^3 \), where \( a \) and \( b \) are non-zero digits and \( n \) is an integer, we can follow these steps:

### Step 1: Understand the left-hand side (LHS)
The expression \( aabb \) represents a four-digit number where:
- The thousands and hundreds digits are both \( a \).
- The tens and units digits are both \( b \).

We can express \( aabb \) mathematically as:
\[ aabb = 1000a + 100a + 10b + b = 1100a + 11b \]

### Step 2: Understand the right-hand side (RHS)
The expression on the right-hand side is \( n^4 - 6n^3 \). We can factor this expression:
\[ n^4 - 6n^3 = n^3(n - 6) \]

### Step 3: Set up the equation
Now we equate the two sides:
\[ 1100a + 11b = n^3(n - 6) \]

### Step 4: Analyze the equation
Since \( 1100a + 11b \) is divisible by 11, we can conclude that \( n^3(n - 6) \) must also be divisible by 11. Since 11 is a prime number, either \( n \) or \( n - 6 \) must be divisible by 11.

### Step 5: Consider possible values for \( n \)
1. **Case 1:** Let \( n = 11 \)
\[ n^3(n - 6) = 11^3(11 - 6) = 1331 \times 5 = 6655 \]

2. **Case 2:** Let \( n - 6 = 11 \) (which means \( n = 17 \))
\[ n^3(n - 6) = 17^3(17 - 6) = 4913 \times 11 = 54043 \]
This is a five-digit number, so we discard this case.

### Step 6: Find \( a \) and \( b \)
From Case 1, we found:
\[ aabb = 6655 \]
This means:
- \( a = 6 \)
- \( b = 5 \)

### Step 7: Conclusion
The only solution to the equation \( aabb = n^4 - 6n^3 \) is:
\[ (a, b) = (6, 5) \]

### Final Answer
The final answer is:
\[ \text{The solution is } (a, b) = (6, 5). \]