N is a positive integer such that N² is expressible as the difference of two consecutive perfect cubes, and 2N + 79 is a perfect square.

Determine the maximum value of N.

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

Posted by broll on 2011-05-09 06:35:48