If n^2 is expressible as the difference of two consecutive perfect cubes, then 2n-1 is a square, as was proved elsewhere, see Short and sweet
Now we have 2n-1=a^2, 2n+79=b^2; a^2=b^2-80, with exactly two solutions: {a,b,n} {1,9,1} and {19,21,181}. So the latter is the maximal value of N.
QED
Edited on May 9, 2011, 6:41 am
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