Posted by Matt on Wednesday, July 15, 2009 at 11:27pm.
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
Nice Answer, But Please Try To Use (Mod)
That Way Is Easier
Hey, Your ANSWER is corrupt, cause it doesnt really explain anything! Try to make it more clear.
SY
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