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

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SUMMARY

The discussion focuses on proving that if \(x^3 + 2y^3 + 4z^3 = 0\), then \(x\), \(y\), and \(z\) must all be even integers. Participants highlight that the cube of any even number is even, while the cube of an odd number is odd. It is established that since \(2y^3 + 4z^3\) is even, \(x^3\) must also be even to satisfy the equation. Furthermore, it is noted that the cube of an even number is divisible by 8, reinforcing the conclusion that \(y\) and \(z\) must also be even.

PREREQUISITES
  • Understanding of integer properties and definitions, particularly even and odd numbers.
  • Familiarity with algebraic manipulation and cubic equations.
  • Knowledge of divisibility rules, specifically regarding powers of integers.
  • Basic experience with mathematical proofs and logical reasoning.
NEXT STEPS
  • Study the properties of even and odd integers in depth.
  • Learn about algebraic identities and their applications in proofs.
  • Explore the concept of divisibility, particularly with powers of integers.
  • Practice constructing mathematical proofs, focusing on integer equations.
USEFUL FOR

Mathematics students, educators, and anyone interested in number theory or algebraic proofs will benefit from this discussion.

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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