Ady,

I nearly follow, but not quite. Using Excel, plainly fractions of the form '(9n-1)/3' will work.

Substituting: 3((9n-1)/3)^3 +10((9n-1)/3)^2 - 3((9n-1)/3) = 81n^3+63n^2-26n+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 2013-03-17 06:18:33