Solving Integer Questions: Showing x, y, and z are Even

[garyljc]

i've come across a question thatr reads
x^3 + 2y^3 + 4z^3 =0
show that x y z are all even

part 2 requires to show that there are no such intergers

i have no idea at all how to show something is even
can anyone help please thanks

 

[HallsofIvy]

It helps if you know that the cube of any even number is even and the cube of any odd number is odd! Obviously, for any y and z, 2y^3+ 4z^3= 2(y^3+ 2z^3) is even. In order that x³ cancel that, x³ must be even.

But you can say more. The cube of an even number is divisible by 8: (2n)³= 8n³. So if x is even, what about 2(y³+ 2z³)? And then what about (y³+ 2z³)?

 

[garyljc]

thanks for the hints !

 

[HallsofIvy]

Oh, and you show something is even by showing it satisfies the definition of "even": it is equal to 2n for some integer n.

 

[garyljc]

got it ! =)