Why three-digit integers only? (Part-2)

If \(A\) and \(B\) are three-digit positive integers, let \(A \text{ # } B\) denote the six-digit integer formed by placing \(A\) and \(B\) side by side. Find the SUM of all the possible values of \(A\) and \(B\) such that:

\[A, \quad B, \quad B-A, \quad A \text{ # } B, \quad \dfrac{A\text{ # }B}{B}\]

are all integer squares.

Note that \(A, B \geq 100\).

If you want to solve it's sister problem, an easy version, try this: Why three-digit integers only?.

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