Prove Difference of Squares of Odd #s is Multiple of 8

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Homework Help Overview

The problem involves proving that the difference between the squares of any two odd numbers is a multiple of 8. The discussion centers around the algebraic manipulation of expressions involving odd integers and their properties.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the algebraic expression derived from the difference of squares and question whether showing it is a multiple of 4 is sufficient to conclude it is a multiple of 8. Some suggest reasoning based on the parity of the integers involved.

Discussion Status

The discussion is ongoing, with participants exploring different reasoning paths and clarifying concepts related to even and odd integers. Some guidance has been offered regarding the properties of even numbers, but there is no explicit consensus on the proof method yet.

Contextual Notes

Participants express uncertainty about the implications of proving the expression is a multiple of 4 and whether that directly leads to a conclusion about multiples of 8. There is also a mention of confusion regarding the reasoning process.

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hi, i have nearly done this problem but made a mistake somewhere, hope you can help, thnx

Homework Statement



Prove that the difference between the equares of any two odd numbers is a multiple of 8.

Homework Equations



n/a

The Attempt at a Solution



where r is an integer, and n is an integer:

(2r-1)^2 - (2n-1)^2
4r^2 - 4r + 1 - 4n^2 + 4n - 1
4r^2 - 4r - 4n^2 + 4n
4(r^2 - r - n^2 + n)

now, that would show it to be multiple of 4, does this then suffice for proof for a multiple of 8? (8 a multiple of 4)

thnx, just need a quick confirmation of this..

cheers
 
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You can reason this way:
1) If both r and n are even, then each of r^2-n^2 and n-r must differ by a multiple of two, so their sum does too. The overall quantity is thus a multiple of eight.

You can do the other two cases...
 
Or you can just show r^2-r is even for ANY r (so clearly so is n^2-n).
 
marcusl said:
You can reason this way:
1) If both r and n are even, then each of r^2-n^2 and n-r must differ by a multiple of two, so their sum does too. The overall quantity is thus a multiple of eight.

You can do the other two cases...

soz, i don't mean to sound super noobish, but i don't quite understand how that works :S soz

cna't someone please elabourate? thnx
 
if r an n are both even , then r^2 and n^2 are even.
that means r^2-n^2 is even, now since r and n are even then r-n is even.
so (r^2 - r - n^2 + n) is even.
...
 
o rite, i see that, but where does it go from there?
 
"now, that would show it to be multiple of 4, does this then suffice for proof for a multiple of 8? (8 a multiple of 4)"

Just for illustration purposes, 12 is a multiple of 4, because 4(3) = 12
Does that mean 12 is a multiple of 8?
 
drpizza said:
"now, that would show it to be multiple of 4, does this then suffice for proof for a multiple of 8? (8 a multiple of 4)"

Just for illustration purposes, 12 is a multiple of 4, because 4(3) = 12
Does that mean 12 is a multiple of 8?

o yeh, duh! stupid me, lol, i was getting in a muddle
 

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