(In reply to
allof tem  spoiler by Ady TZIDON)
Ady,
I nearly follow, but not quite. Using Excel, plainly fractions of the form '(9n1)/3' will work.
Substituting: 3((9n1)/3)^3 +10((9n1)/3)^2  3((9n1)/3) = 81n^3+63n^226n+2, with no fractional part. So the substitution works as advertised (it is because the 1/9 in the expansion of the cubed term cancels aganst the 10/9 in the expansion of the squared term to produce an integral result).
So the fractions would be 8/3,17/3, 26/3 etc. with negative terms like 1/3, as you mention, but 3 not 9 as the denominator.
Edited on March 17, 2013, 6:30 am

Posted by broll
on 20130317 06:18:33 