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Little cubes (Posted on 2010-02-27) |
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The difference of the cubes of two non-negative successive integers is a square of an integer.
Find few possible pairs.
How about a general formula or evaluation algorithm
for a(n)?
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Submitted by Ady TZIDON
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Rating: 4.0000 (2 votes)
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Solution:
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(Hide)
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Some couples (a,b) for (a+1)^3-a^3=b^2 :
(0, 1), (7, 13), (104, 181),...
To get more, use the recurrence relations :
a[n] = 14a[n-1] - a[n-2] + 6 and
b[n] = 14b[n-1] - b[n-2]
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