Division Algorithm: Proving 24 Does Not Divide a² - 1

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Homework Help Overview

The discussion revolves around proving that if an integer \( a \) is not divisible by 2 or 3, then 24 does not divide \( a^2 - 1 \). Participants are exploring the implications of the conditions on \( a \) and how they relate to divisibility by 24.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to analyze the expression \( a^2 - 1 \) by considering cases based on the form of \( a \) when it is not divisible by 2 or 3. Some participants question the validity of certain expressions and the implications of divisibility by 6.

Discussion Status

Participants are actively engaging with the problem, offering insights into the structure of \( a^2 - 1 \) based on different forms of \( a \). There is a recognition of the need to show that certain factors cannot both be odd, indicating a productive line of reasoning. However, there is no explicit consensus on the approach being taken.

Contextual Notes

There is a focus on the specific conditions that \( a \) must satisfy, and some participants express confusion about how to relate these conditions to the divisibility of \( a^2 - 1 \). The discussion reflects an ongoing exploration of the problem without a complete resolution.

MAGICMATHS
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1. My difficulty is to show that if a is an integer such that 2 does not divide a and 3 does not divide a then 24 does not divide a squared minus 1

2.Is there any equation which helps?

3. My idea is that it has to be an integer such that 6 does not divide a...therefore i have to show that 6a,6a+1,...6a+5 converted to asquared minus 1 are not divisible by 24...pls can anyone suggest if I am solving it correctly.
 
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Well, not "6a" because that would be divisible by both 2 and 3. And you don't mean "6a+ ...". If "2 does not divide a and 3 does not divide a" then either a= 6k+ 1 or a= 6k+ 5 for some integer k.

Now, if a=6k+ 1, [itex]a^2- 1= 36k^2+ 12k+1-1= 36k^2- 12k= 12k(3k- 1)[/itex]. That will be divisible by 3 if either k or 3k-1 is even- and it is easy to see that they can't both be odd.

If a= 6k+ 5, [itex]a^2-1= 36k^2+ 60k+ 25- 1= 36k^2+ 60k+ 24= 12(3k^2+ 5k+ 2)= 12(3k+ 2)(k+ 1)[/itex]. Now it is only necessary to show that 3k+2 and k+ 1 can't both be odd.
 
how can i explain that 6k is divisible by 6 when if i change it to asquared - 1 (pls refer to previous message) it becomes 36ksquared - 1...thats not easy to explain as divisible by 6!:(
 
The hypothesis is "a is an integer such that 2 does not divide a and 3 does not divide a" so [itex](6k)^2- 1[/itex] is not relevant.
 

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