# 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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