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

  • Thread starter Thread starter Trail_Builder
  • Start date Start date
AI Thread Summary
The discussion revolves around proving that the difference between the squares of any two odd numbers is a multiple of 8. The initial attempt simplifies the expression to show it is a multiple of 4 but questions whether this suffices for a multiple of 8. Participants clarify that if both integers are even, the resulting expressions will also yield even results, leading to a conclusion that the overall difference is indeed a multiple of 8. The conversation highlights the importance of understanding the relationship between multiples of 4 and 8 in this context. Ultimately, the proof confirms that the difference of squares of odd numbers is a multiple of 8.
Trail_Builder
Messages
148
Reaction score
0
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
 
Physics news on Phys.org
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
 
Back
Top