Prove that if a and b are both odd integer

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Homework Help Overview

The discussion revolves around proving a statement regarding odd integers, specifically that if a and b are both odd integers, then 16 divides (a^2 + b^2 - 2). Participants are exploring the implications of this statement within the context of number theory.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to express a and b in terms of integers m and n, leading to a partial proof that shows divisibility by 8 but seeks further guidance on how to proceed. Some participants question the validity of the original statement by providing counterexamples, suggesting that the assertion may be incorrect. Others explore the implications of the expression a^2 + b^2 - 2 in relation to divisibility by 16.

Discussion Status

The discussion is ongoing, with participants actively questioning the original problem statement and providing counterexamples. There is no consensus on the validity of the claim, and multiple interpretations of the problem are being explored.

Contextual Notes

Some participants note specific counterexamples that challenge the original assertion, indicating potential issues with the problem as stated. The discussion reflects uncertainty about the divisibility conditions and the implications of the arithmetic involved.

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Homework Statement



prove that if a and b are both odd integer, then [tex]16|(a^2+b^2-2)[/tex]

Homework Equations



n/a

The Attempt at a Solution



let [tex]a=2m+1 \ and \ b=2n+1[/tex], then [tex]a^2+b^2-2=4(m(m+1)+n(n+1))[/tex] so its divisible by 4, and also divisible 8 since [tex]m(m+1) \ and \ n(n+1)[/tex] are even.
so i only prove [tex]8|(a^2+b^2-2)[/tex], then how to continue? clue please T_T
 
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hey I've got counter example, a=1 and b=3, so the question is wrong?
 


16\8=0 or am i wrong?
 


A thought..Let a^2 +b^2-2<> 16k. a^2+b^2<>16k+2 => a^2+b^2<>2m => odd +odd<> even which is wrong. (<> different from)
 


annoymage said:
hey I've got counter example, a=1 and b=3, so the question is wrong?

3^2+7^2-2=56. That isn't divisible by 16 either. Yes, there is something wrong with the question.
 

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