# Completing Cube

Prove or disprove that There exists a positive integer M, such that 1^3+2^3+…+n^3=M^3, where n>1.

Thank you for helping.

Simon YAU.

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"There exists a positive integer M,

such that 1³ + 2³ + … + n³ = M³, where n > 1."

First of all, we know a common result about the sum of cubes:

1³ + 2³ + … + n³ = [ n(n+1)/2 ]²

Therefore, the statement is equivalent to

"There exists a positive integer M,

such that [ n(n+1)/2 ]² = M³, where n > 1."

If a number is both a perfect square and a perfect cube, it must be a perfect sixth power.

That is, [ n(n+1)/2 ]² = M³ = A⁶ where A is a positive integer.

In other words, we need that

n(n+1)/2 = A³ is a perfect cube, and M = A² is a perfect square.

(Actually, the only triangular number n(n+1)/2 which is also a perfect cube is only 1.)

Consider

n(n+1)/2 = A³

n² + n = 2A³

4n² + 4n = 8A³

4n² + 4n + 1 = 8A³ + 1

(2n + 1)² = (2A)³ + 1

(2n + 1)² - (2A)³ = 1 ...(✿)

Among all integers p and q such that p² - q³ = 1,

the ONLY solution is p = 3 and q = 2.

This is called the Catalan's conjecture (now a theorem).

In other words, the only solution to (✿) is that

2n + 1 = 3 and 2A = 2M² = 2

That is, n = 1 and M = 1.

Therefore, the ONLY possible case is n = 1.

That means, there does not exists a positive integer M

such that 1³ + 2³ + … + n³ = M³, where n > 1.

References:

http://en.wikipedia.org/wiki/Squared_triangular_nu...

http://mathworld.wolfram.com/CubicTriangularNumber...

http://mathworld.wolfram.com/CatalansConjecture.ht...

Basic references:

http://www.themathpage.com/arith/appendix.htm

http://mathcentral.uregina.ca/QQ/database/QQ.09.06...

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