You are not logged in.
Pages: 1
Prove that the square of an odd number is of the form 8q + 1.
Let m = an odd number
All numbers can be represented in the form 4k ± 1.
Hence, m = 4k ± 1.
m² = (4k ± 1)² = 16k² ± 8k + 1 = 8k(2k ± 1) + 1
But what if you choose 2k ± 1? Won't that give the same answer?
m² = (2k ± 1)² = 4k² ± 4k + 1 = 4k(k ± 1) + 1
Huh? Now it's only of the form 4q + 1???
Offline
Yes, but in that case q is always even. If k is even then k(k±1) is even, but if k is odd then k±1 is even and so k(k±1) is still even. You can take a factor of 2 out of q safe in the knowledge that it will always remain an integer.
Why did the vector cross the road?
It wanted to be normal.
Offline
In general, I would stay away from the ± and just use the standard definition of odd:
2k - 1
± will probably make the problem a lot harder than it actually is.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
Offline
n^2= (2k+1)^2= 4k^2+ 4k+ 1= 4(k)(k+1)+1.
If k is even, k= 2m, then this is 4(2m)(2m+1)+1= 8(m)(2m+1)+ 1, a multiple of 8 plus 1.
If k is odd, k= 2m+1, then k+1= 2m+1+1= 2(m+1) so this is 4(2m+1)(2)(m+1)+1 = 8(2m+ 1)(m+1)+ 1, a multiple of 8 plus 1.
(I'm posting this because I started to ask HOW would you use n= 2k+1 to prove that n^2 is a multiple of 8 plus 1, then saw how while I was typing my question!)
Another way of proving the result is by mathematical induction (not on n but on k).
Offline
Pages: 1