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| | Unit Fraction Partitions |
 | | If k=1 then the only integer solutions (a,b,c) correspond to the positive integer solutions (X,Y,Z) of (3), namely (1,2,5) (1,3,3) (2,2,2) For k=2 we can accept half-integer solutions of (3), and so on. |
 | | Another approach is based on the fact that for any positive integer k there exists a least integer D(k) such that the partitions of 1/k into three unit fractions correspond to the partitions of D(k)/k into three divisors of D(k). |
 | | Here the only effect of k is to determine which of the rational solutions of (3) give integer values of a,b,c according to a = (X+1)k b = (Y+1)k c = (Z+1)k In other words, k just "scales up" the rational solutions of (2), resulting in a certain number of them being integers. |
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