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 Square and Consecutive Cubes (Posted on 2009-12-17)
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.

 No Solution Yet Submitted by K Sengupta No Rating

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 re: A start! | Comment 2 of 4 |
(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±√(b²-4ac)              -3±√(9-12(1-N²))
----------------- ====> --------------------- ====>
2a                                 6

A = -0.5 ± √(N²/3-1/12) ...

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**2-1)/12)**.5-.5
if ((a+1)**3-a**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 2009-12-17 16:21:44

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