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

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Homework Help Overview

The discussion revolves around a mathematical problem involving the equation x3 + 2y3 + 4z3 = 0, where the goal is to demonstrate that the integers x, y, and z are all even. Additionally, there is a second part of the problem that requires showing the non-existence of such integers.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the properties of even and odd numbers, particularly focusing on the implications of cubing even and odd integers. There is an exploration of how the evenness of x, y, and z can be inferred from the structure of the equation.

Discussion Status

Some participants have provided hints regarding the properties of even and odd numbers, suggesting that the discussion is moving towards a clearer understanding of the problem. There is acknowledgment of the definition of even numbers, which may guide further reasoning.

Contextual Notes

Participants express uncertainty about how to demonstrate that certain integers are even, indicating a potential gap in foundational knowledge or understanding of the definitions involved.

garyljc
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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
 
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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)?
 
thanks for the hints !
 
Oh, and you show something is even by showing it satisfies the definition of "even": it is equal to 2n for some integer n.
 
got it ! =)
 

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