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November 23, 2014

November 23, 2014

Posted by **m** on Thursday, July 23, 2009 at 7:18am.

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

The question is 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.

Thanks> please explain in full steps and in the most logical order

Thanks once again

- math -
**MathMate**, Thursday, July 23, 2009 at 8:09amThe proof does look clumsy. It is however a little easier to follow if the exponents are displayed correctly. See the original version here:

http://www.jiskha.com/display.cgi?id=1247714839

- math -
**Jon Zhang**, Monday, August 17, 2009 at 5:57amDude, it's going to be due in like three days. Do it yourself.

- math -
**Sean**, Tuesday, August 18, 2009 at 6:41amit's due in two days, lol

SY

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