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Cubic Consecutive Conundrum (Posted on 2024-12-05)

It is given that 181² can be written as the difference of the cubes of two consecutive positive integers. Find the sum of these two integers.

No Solution Yet Submitted by Danish Ahmed Khan

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Solution

| Comment 1 of 5

(n+1)^3 - n^3 = 181^2

3n^2 + 3n + 1 = 32761

3n^2 + 3n = 32760

4n^2 + 4n = 43680

4n^2 + 4n + 1 = 43681

(2n+1)^2 = 209^2

n+1 + n = 209

Posted by Brian Smith on 2024-12-05 13:48:59

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