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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 2 of 5 |

let a-b = 1

a^3 - b^3 = (a-b)(a^2 + ab + b^2) = 181^2

(a^2 + ab + b^2) = 181^2

(a^2 + a(a-1) + (a-1)^2) = 181^2

(a^2 + a^2 - a + a^2 - 2a + 1) = 181^2

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

3a^2 - 3a + (1-181^2) = 0

3a^2 - 3a - (180)(182) = 0

a^2 - a - (60)(182) = 0

a = {+105 or -104}, but we need positive, so

a = 105

b = a-1 = 104

a+b = 209

check:

105^3 - 104^3 = 32761

181^2 = 32761

Posted by Larry on 2024-12-05 16:09:14

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