(In reply to
A start! by rod hines)
For N² to be the difference between two consecutive cubes (A, and A+1), N² = 3A²+3A+1. Subtracting N² from both sides to create a quadratic:
b±√(b²4ac) 3±√(912(1N²))
 ====>  ====>
2a 6
A = 0.5 ± √(N²/31/12) ...
Running through a quick program to find all possible values of N (up to 100,000,000) that lead to an integer result, we get:
1,13,181,2521,35113,489061,6811741,94875313
Running the following code:
def square_and_consecutive_cubes():
possible=[1,13,181,2521,35113,489061,6811741,94875313]
for n in possible:
a=((4*n**21)/12)**.5.5
if ((a+1)**3a**3==n**2)*is_perfect_square(2*n+79)==1:
print(n,a,a+1)
Gets us two solutions for N<100,000,000:
N=1, A=0 and N=181, A=104. Not that this proves anything, but I do believe that N=181 is likely the largest value such that N² = (A+1)³  A³ and 2N + 79 is a perfect square.

Posted by Justin
on 20091217 16:21:44 