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@ Number Theory

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1 @ Number Theory on Sun Mar 07, 2010 4:30 am

Lucifer


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Can 6*n^2 + 12*n + 8 be a cube of a natural number for any natural number n ? Give proof.

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2 Re: @ Number Theory on Sat Mar 13, 2010 6:36 am

can somebody prove that (k+1)^3 -k^3 is not a cube! That will help me solve this prob.

because 6*n^2 + 12*n + 8 = n^3 + 6*n^2 + 12*n + 8 - n^3 = (n+2)^3 - n^3 = 2((n+2)^2 + n(n+2) + n^2). Now substituting n = 2k We get 2^3 * ( 3*k^2 + 3*k +1). for this to be a cube 3*k^2 + 3*k +1 must be a cube!

3*k^2 + 3*k +1 = (k+1)^3 - k^3.

I am stuck here. All I have to prove is that (k+1)^3 - k^3 is not the cube of a natural number. Any help?

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3 @indranil on Sat Mar 13, 2010 10:59 pm

Lucifer


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6*n^2 + 12*n + 8 = n^3 + 6*n^2 + 12*n + 8 - n^3 = (n+2)^3 - n^3,
now we need to find k such that,
k^3 = (n+2)^3 - n^3,
=> (n+2)^3 = n^3 + k^3,

Well by using Fermat's last theorem, it is obvious that there is no such k.

Fermat's last theorem:
There are no three positive integers a, b, and c that can satisfy the equation a^n + b^n = c^n for any integer value of n>2.

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4 Re: @ Number Theory on Wed Mar 17, 2010 10:24 pm

damn right !!!

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