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: Express \( aabb \) in terms of \( a \) and \( b \)

The number \( aabb \) can be expressed as:
\[ aabb = 1000a + 100a + 10b + b = 1100a + 11b \]

### Step 2: Set up the equation

Now we can equate this to the right-hand side of the original equation:
\[ 1100a + 11b = n^4 - 6n^3 \]

### Step 3: Factor the right-hand side

We can factor the right-hand side:
\[ n^4 - 6n^3 = n^3(n - 6) \]

### Step 4: Analyze the equation

Now we have:
\[ 1100a + 11b = n^3(n - 6) \]

### Step 5: Simplify the equation

We can simplify the left-hand side:
\[ 11(100a + b) = n^3(n - 6) \]

### Step 6: Determine the divisibility condition

Since \( 11 \) is a common factor, we can conclude that \( n^3(n - 6) \) must also be divisible by \( 11 \). This means either \( n \) or \( n - 6 \) must be divisible by \( 11 \).

### Step 7: Check possible values of \( n \)

Let’s check possible integer values for \( n \) that are divisible by \( 11 \) or \( n - 6 \) is divisible by \( 11 \).

1. **Case 1:** \( n = 11k \) for some integer \( k \)
2. **Case 2:** \( n = 11k + 6 \) for some integer \( k \)

### Step 8: Calculate \( n^3(n - 6) \) for small values of \( n \)

Let’s try small integer values for \( n \):

- For \( n = 0 \):
\[ n^3(n - 6) = 0 \]

- For \( n = 1 \):
\[ n^3(n - 6) = 1^3(1 - 6) = -5 \]

- For \( n = 2 \):
\[ n^3(n - 6) = 2^3(2 - 6) = 8 \times -4 = -32 \]

- For \( n = 3 \):
\[ n^3(n - 6) = 3^3(3 - 6) = 27 \times -3 = -81 \]

- For \( n = 4 \):
\[ n^3(n - 6) = 4^3(4 - 6) = 64 \times -2 = -128 \]

- For \( n = 5 \):
\[ n^3(n - 6) = 5^3(5 - 6) = 125 \times -1 = -125 \]

- For \( n = 6 \):
\[ n^3(n - 6) = 6^3(6 - 6) = 216 \times 0 = 0 \]

- For \( n = 7 \):
\[ n^3(n - 6) = 7^3(7 - 6) = 343 \times 1 = 343 \]

- For \( n = 8 \):
\[ n^3(n - 6) = 8^3(8 - 6) = 512 \times 2 = 1024 \]

- For \( n = 9 \):
\[ n^3(n - 6) = 9^3(9 - 6) = 729 \times 3 = 2187 \]

- For \( n = 10 \):
\[ n^3(n - 6) = 10^3(10 - 6) = 1000 \times 4 = 4000 \]

- For \( n = 11 \):
\[ n^3(n - 6) = 11^3(11 - 6) = 1331 \times 5 = 6655 \]

### Step 9: Check for valid \( a \) and \( b \)

Now we need to check which of these results can be expressed in the form \( 1100a + 11b \) where \( a \) and \( b \) are non-zero digits (1 to 9).

- For \( n = 7 \):
\[ 343 = 1100a + 11b \]
This does not yield valid digits.

- For \( n = 8 \):
\[ 1024 = 1100a + 11b \]
This does not yield valid digits.

- For \( n = 9 \):
\[ 2187 = 1100a + 11b \]
This does not yield valid digits.

- For \( n = 10 \):
\[ 4000 = 1100a + 11b \]
This does not yield valid digits.

- For \( n = 11 \):
\[ 6655 = 1100a + 11b \]
This does not yield valid digits.

### Conclusion

After checking all possible values of \( n \) up to \( 11 \), we find that there are no valid solutions for \( a \) and \( b \) such that \( aabb = n^4 - 6n^3 \) with \( a \) and \( b \) being non-zero digits.