Let n be any natural number whose square is also the difference between two consecutive cubes.
Prove that 2n1 is also a square.
3a^2 + 3a + 1 = n^2
If you multiply this equation by 4 and subtract 1, you can recast it as:
3(2a+1)^2 = (2n+1)(2n1)
3 can factor only 1 of the factors of the RHS, which gives us 2 cases:
1) 2n+1 = p^2
2n1 = 3q^2
2) 2n+1 = 3p^2
2n1 = q^2
1) implies 3q^2 + 2 = p^2, which is impossible as no square = 2mod3.
So 2n1 = q^2, which answers the problem.
Going further and setting q=2r+1 gives n = r^2 + (r+1)^2.
So you have the neat conclusion that when the difference of 2 consecutive cubes is a square, the root of that square is the sum of 2 consecutive squares.

Posted by xdog
on 20110223 18:48:46 