100 is a square which is also a sum of 4 consecutive cubes.
Are there any others?
Start with Jer's factoring: y^2 = 2(2x+3)(x²+3x+6).
If d = greatest common divisor of (2x+3) and (x²+3x+6), it also divides 4 times the second expression minus the square of the first expression = 15. So d=1,3,5,15. The following handles d=1.
Since (2x+3) can't be twice anything, set
(2x+3) = a^2 and (x^2+3x+6) = 2b^2
Solve the first for x and substitute in the second to get
(a^4+15) = 8b^2 = 2c^2 for c = 2b
So (2c^2 - 15) is the square of a integer.
If c = 0 or 5 mod 10 then (2c^2 - 15) = 85 or 35 mod 100 which is impossible.
Other values of c mod 10 give (2c^2 - 15) = 3 or 7 mod 10 which is also impossible for squares.
Posted by xdog
on 2013-06-21 10:59:46