Direct proof using definiton of even

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SUMMARY

The discussion focuses on proving that for all integers n and m, if n - m is even, then n³ - m³ is also even. The proof begins with the definition of even integers, expressed as n = 2k. The user initially attempts to manipulate the expression n³ - m³ using algebra but finds it unconvincing. Ultimately, they realize that applying the identity x³ - y³ = (x - y)(x² + xy + y²) effectively demonstrates that n³ - m³ is even, confirming the proof.

PREREQUISITES
  • Understanding of integer properties and definitions, particularly the definition of even integers.
  • Familiarity with algebraic manipulation and polynomial identities.
  • Knowledge of basic proof techniques in mathematics.
  • Experience with cubic functions and their properties.
NEXT STEPS
  • Study the algebraic identity x³ - y³ = (x - y)(x² + xy + y²) in depth.
  • Explore additional properties of even and odd integers in number theory.
  • Practice constructing proofs using definitions and algebraic identities.
  • Learn about mathematical induction as a proof technique for integer properties.
USEFUL FOR

Students in mathematics, particularly those studying number theory or proof techniques, as well as educators looking for examples of proofs involving even and odd integers.

cmajor47
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Homework Statement


Prove that for all integers n and m, if n-m is even then n3-m3 is even.


Homework Equations


Definition of even: n=2k


The Attempt at a Solution


Proof: Let n, m [tex]\in[/tex] Z such that n-m=2k
n-m=2k
n=2k+m
m=-2k+n
n3-m3=(2k+m)3-(-2k+n)3

I did all of this algebra out but I didn't think that it worked in showing that n3-m3 is even. Am I doing the proof wrong?
 
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How about using the fact that x3- y3= (x- y)(x2+ xy+ y2)?
 
Last edited by a moderator:
Thanks so much, I realized how to do the proof with that help.
 

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