Does That 2 Really Count?

Number Theory Level 1

$x$ and $y$ are two odd positive integers such that $x-y=2$ .
The number $x^3-y^3$ is $\text{\_\_\_\_\_\_\_\_\_\_\_\_\_} .$

divisible by both 2 and 3 divisible by 2 but not by 3 divisible by 3 but not by 2 divisible by neither 2 nor 3

1 solution

Geoff Pilling
Nov 7, 2016

Substituting in $x=y+2$ ...

$x^3 - y^3 = (x-y)(x^2 + xy + y^2) = 2((y+2)^2 + (y+2)y + y^2) = 2(y^2+4y+4 + y^2 + 2y + y^2)$

$x^3 - y^3 = 2(3y^2 + 6y + 4)$

Clearly, this product must divide by $2$ .

Since neither $2$ nor $3y^2 + 6y + 4$ can divide by $3$ , then the total can't divide by $3$ .

Therefore, $x^3 - y^3$ is $\boxed{\text{Divisible by 2 but not by 3}}$