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A Mathematical Challenge:

Humorous illustration about the perfect square number
— Of course, I'm a square.
Problem:

Find a four-digit natural number of the form \( n = \overline{aabb} \) that is a perfect square.

Note: The solution is expected to avoid brute-force methods and instead rely on conceptual analysis before attempting numerical trials.

Solution:

Starting with the number in the form \( n = \overline{aabb}_{10} \), we expand it using positional notation:

\[ \begin{align*} n &= 1000a + 100a + 10b + b \\ &= 1100a + 11b = 11(100a + b) \end{align*} \]

1. Divisibility by 11

To analyse the expression \(100a + b = (a0b)_{10}\), we apply the divisibility rule for 11:

A number is divisible by \(11\) if and only if the difference between the sum of its digits in odd positions and the sum in even positions is a multiple of \(11\).


For the number \((a0b)_{10}\):

  1. Hundreds digit: \(a\) (odd position)
  2. Tens digit: \(0\) (even position)
  3. Units digit: \(b\) (odd position)

Applying the rule:

\[ (a + b) - 0 = a + b \equiv 0 \pmod{11} \]


Since \(a\) and \(b\) are digits (\(1 \leq a \leq 9\), \(0 \leq b \leq 9\)), the only possibility is:

\[ a + b = 11 \quad \Rightarrow \quad b = 11 - a \]

2. Algebraic development

Substituting into the original expression \(n = 11(100a + b)\), we get:

\[ n = 11(100a + 11 - a) = 11(99a + 11) = 11^2(9a + 1) \]

For \(n\) to be a perfect square, \(9a + 1\) must also be a perfect square. That is, there exists a natural number \(x\) such that \(9a + 1 = x^2\).

5. Final Answer

We conclude that:

\[ n = 7744 \]


is the only perfect square of the form \( \overline{aabb} \).

If you’ve followed the reasoning carefully this far, you might enjoy going a bit further and exploring the history and original sources of this problem .