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2-digit Square

Thank you Abdul for this solution.

The 2-digit number is either $65$ or $56$.


Any 2-digit number can be represented as $10a + b$. We need $(10a+b)^2 - (10b+a)^2 = 99a^2 - 99b^2 =9 \times 11 \times (a^2 - b^2)$ to be a square.

This means that $(a^2 - b^2)$ must be 11 and so $(a - b)(a + b) = 11$ making, $a - b = 1$ and $a + b = 11$. This gives $a = 6$, and $b = 5$.

If we find a solution with $a > b$ then, by reversing the digits, we get a second solution.