To prove that there are no square numbers greater than a single digit that have all of their digits the same, we can proceed with a logical reasoning approach combined with a bit of number theory. Let's break down the proof step-by-step:

Step-by-Step Proof:

1. Understanding the Problem:

   • A square number is a number that can be expressed as $n^2$ where $n$ is an integer.
   • A number with all its digits the same is of the form $\underbrace{aaa\ldots a}_{k \text{ digits}}$, where each digit $a$ is the same.
   • We need to prove that there are no square numbers (other than 1, 4, or 9) where every digit is identical.

2. Form of Numbers with Identical Digits:

   • Any number with $k$ identical digits $a$ can be expressed as: $N = a \times \frac{10^k - 1}{9}$ Here, $\frac{10^k - 1}{9}$ is a number with $k$ digits, all of which are 1. For example, for $k = 3$, $\frac{10^3 - 1}{9} = 111$.

3. Possible Values of $a$:

   • Since every digit $a$ must be the same and within the range of digits (0-9), $a$ must be a digit from 1 to 9 (0 is not allowed because 000...0 isn't a valid number in this context).

4. Analyzing the Number $N$:

   • If $N$ is a perfect square, $N = m^2$ for some integer $m$.
   • Now, $N = a \times \frac{10^k - 1}{9} = m^2$.

5. Constraints Imposed by the Form of $N$:

   • The number $\frac{10^k - 1}{9}$ has a specific form depending on $k$, and its properties determine whether $N = a \times \frac{10^k - 1}{9}$ can be a square.

6. Checking for Possible Squares:

   • Notice that $\frac{10^k - 1}{9}$ has the form $111\ldots1$, and it is only divisible by 3 if $k$ is a multiple of 3. This form isn't generally a perfect square. Let's analyze the number in terms of modulo arithmetic.

7. Modulo Analysis:

   • For any $k \geq 2$, let's consider the squares modulo 9:
     • Possible residues of perfect squares modulo 9 are 0, 1, 4, or 7.
     • However, the repeating-digit numbers for $k \geq 2$ are not congruent to 0, 1, 4, or 7 modulo 9 for any $a = 2, 3, \ldots, 9$.

8. Conclusion from Modulo Analysis:

   • Given the values we derived and tested through modulo analysis and the constraints of the repeating numbers, no number of the form $aaa\ldots a$ (with $k > 1$) can be a perfect square.
   • Hence, no square numbers other than 1, 4, or 9 have all their digits the same.

Conclusion:

Therefore, there are no square numbers other than the single-digit numbers $1^2 = 1, 2^2 = 4,$ and $3^2 = 9$ that have all identical digits.

Would you like to see more details or have any questions about this proof?

Related Questions:

1. What are the perfect squares less than 100?
2. Can a number with all identical digits be a cube?
3. How does modulo arithmetic help in number theory proofs?
4. Are there any other forms of numbers with special properties regarding their digits?
5. Can we prove a similar property for numbers in bases other than 10?

Tip: When solving problems involving properties of numbers, modulo arithmetic can be a very powerful tool to quickly check the feasibility of certain properties or constraints.