Monday

September 22, 2014

September 22, 2014

Posted by **m** on Saturday, July 25, 2009 at 6:39am.

prove that a number 10^(3n+1), where n is a positive integer, cannot be represented as the sum of two cubes of positive integers.

with out using this method at all

.................................

We will examine the sum of cubes of two numbers, A aand B. Without losing generality, we will further assume that

A=2nX and

B=2n+kY

where

X is not divisible by 2

n is a positive integer and

k is a non-negative integer.

A3+B3

=(A+B)(A2-AB+B2)

=2n(X + 2kY) 22n(X2 - 2kXY + 22kY²)

=23n(X + 2kY) (X² - 2kXY + 22kY²)

Thus A3+B3 has a factor 23n, but not 23n+1 since X is not divisible by 2.

Since 103n+1 requires a factor of 23n+1, we conclude that it is not possible that

103n+1=A3+B3

dont use this method.........

can you please answer the question in full steps, thanks

- maths -
**Jon Zhan**, Monday, August 3, 2009 at 8:46pmYou Know If Your Are trying To Solve The Whole Maths Enrichment By Yourself, You Should NOT Ask For Help On The Internet. Plus You Still Have about Two Weeks.

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