# maths

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 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

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

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