Let's find all possible solutions for \(A\ \text{and}\ B\).

\(\dfrac1A+\dfrac1B = \dfrac{A+B}{AB}\) which implies that the number is an integer iff \(AB\ |\ A+B\).

Clearly all pairs of solutions where \(A = -B \neq 0\) are valid.

Now if \(A+B \neq 0\), \(AB\ |\ A+B\) implies that \(A \ |\ A+B\implies A \ |\ B\ \text{and}\ B \ |\ A+B\implies B \ |\ A\) which
implies that \(A=B\).

Now \(A^2 \ |\ 2A\implies A \ |\ 2\).

Using the facts above, we know that the only possible pairs of solutions \((A,B)\) (when \(A+B \neq 0\) ) are:
\((-1,-1),(1,1),(-2,-2),(2,2)\).

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TopNewestLet's find all possible solutions for \(A\ \text{and}\ B\).

\(\dfrac1A+\dfrac1B = \dfrac{A+B}{AB}\) which implies that the number is an integer iff \(AB\ |\ A+B\).

Clearly all pairs of solutions where \(A = -B \neq 0\) are valid.

Now if \(A+B \neq 0\), \(AB\ |\ A+B\) implies that \(A \ |\ A+B\implies A \ |\ B\ \text{and}\ B \ |\ A+B\implies B \ |\ A\) which implies that \(A=B\).

Now \(A^2 \ |\ 2A\implies A \ |\ 2\).

Using the facts above, we know that the only possible pairs of solutions \((A,B)\) (when \(A+B \neq 0\) ) are:

\((-1,-1),(1,1),(-2,-2),(2,2)\).

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I know possible positive values a (1, 1) and (2,2)

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