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How to find ALL pairs of positive integers $(x,y)$ such that the difference in their squares is a perfect cube and the difference in their cubes is a perfect square.
i.e., Positive integers $(x,y)$ such that $x^2-y^2=a^3$ and $x^3-y^3=b^2$ for integral $a, b$?

Finding infinite number of pairs is no problem, as in:

$( 2^{6j+1} \cdot 3^{6k} \cdot 5 , 2^{6j+1} \cdot 3^{6k+1} )$ for any integral $j,k \geq 0$

But how would you determine the exhaustive list?

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    Your example seems to arise from positive integers $r,s$ such that $(r^3 - s^3)/(r^2 - s^2)$ is a perfect square, as appropriate common multiples of $r,s$ then give $x,y$ to satisfy your equations. E.g. $(5^3 - 3^3)/(5^2 - 3^2) = 49$.2011-07-23
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    @hardmath: That is true for my example, but my example does not account for all possibilities and in others that does not hold; e.g., for pairs of the form: $( 2 \cdot 11^{(6k+2)} \cdot 37 , 11^{(6k+2)} \cdot 47 )$2011-07-23

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