# math

posted by on .

can you answer this question in a different and more logical way than this method below:

we will examine the sum of cubes of two numbers, A and 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 - ,

The 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 - ,

Dude, it's going to be due in like three days. Do it yourself.

• math - ,

it's due in two days, lol

SY