How to prove that no square number (besides single digit squares) do not have all the same digits. 

To prove that no square number (besides single-digit squares) has all the same digits, we'll follow a structured argument that leverages basic number theory and properties of digits in numbers.

### **Proof Outline:**

1. **List of Single-Digit Square Numbers:**
   - The single-digit squares are: \( 1^2 = 1 \), \( 2^2 = 4 \), \( 3^2 = 9 \), \( 4^2 = 16 \), \( 5^2 = 25 \), \( 6^2 = 36 \), \( 7^2 = 49 \), \( 8^2 = 64 \), and \( 9^2 = 81 \).
   - Among these, only the square numbers \( 1 \), \( 4 \), and \( 9 \) are single digits and trivially have the same digits (since they are just one digit).

2. **Considering Multi-Digit Square Numbers:**
   - Suppose that there is a multi-digit square number, say \( n^2 \), where all digits are the same.
   - Such a number can be written as \( d \times (111\ldots1) \), where \( d \) is the repeated digit (1 through 9), and the number \( 111\ldots1 \) has \( k \) digits, denoted as \( N_k \).
     - Example: \( 111 \ldots 1 \) with \( k \) digits is the number \( \frac{10^k - 1}{9} \).

3. **Form of Square Numbers:**
   - \( n^2 = d \times N_k \).
   - \( N_k \) can be expressed as \( \frac{10^k - 1}{9} \).
   - Therefore, \( n^2 = d \times \frac{10^k - 1}{9} \), or equivalently, \( 9n^2 = d \times (10^k - 1) \).

4. **Properties of \( 10^k - 1 \):**
   - The number \( 10^k - 1 \) is divisible by 9 but leaves a remainder when divided by any number larger than 9 (except for powers of 10 minus 1).
   - For \( n^2 \) to be equal to \( d \times \frac{10^k - 1}{9} \), \( n^2 \) must be divisible by \( 10^k - 1 \).
   - \( 10^k - 1 \) grows exponentially with \( k \), so for large \( k \), \( n^2 \) (being a polynomial of degree 2) cannot grow fast enough to be equal to \( d \times \frac{10^k - 1}{9} \).

5. **Contradiction in Growth Rates:**
   - The exponential growth rate of \( 10^k - 1 \) and the quadratic growth rate of \( n^2 \) make it impossible for \( n^2 = d \times \frac{10^k - 1}{9} \) for sufficiently large \( k \), except for small cases like single-digit numbers.
   - Therefore, for any multi-digit square number, all digits cannot be the same.

6. **Conclusion:**
   - The only squares with all digits the same are those with a single digit (e.g., \( 1^2 = 1 \), \( 2^2 = 4 \), \( 3^2 = 9 \)). Any multi-digit square will have varying digits.

Would you like more details or further clarification on any part of this proof? 

Here are five related questions:
1. How do we calculate the growth rate of numbers like \( 10^k - 1 \)?
2. What other numbers besides squares might have all the same digits?
3. How can we extend this proof to other powers, like cubes?
4. What are other examples of squares with interesting digit properties?
5. What are some famous problems involving square numbers?

**Tip:** When tackling number theory proofs, it’s essential to focus on the properties of the numbers involved, especially how they grow and their divisibility.

Explore a mathematical proof that demonstrates why no multi-digit square numbers (except single-digit squares) have all the same digits. This detailed analysis uses number theory and properties of digits in numbers. Suitable for advanced high school students and enthusiasts of number theory.

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