prove that the difference of the squares of any two odd natural numbers is divisible by four?

Question

can anybody help me with this math proof . i would really appreciate it.

Answer 1

The general odd number is given by (2*n+1) where n is any integer. Take two arbitrary odd numbers, (2n1+1) and (2n2+1). The difference of their squares is

(2n1+1)^2-(2n2+1)^2
4n1^2+4n1+1 - 4n2^2-4n2-1
4n1^2+2n1-4n2^2-4n2

The latter expression is always divisible by 4 regardless of n1 or n2

Answer 2

Not sure how to do it in proof form but go through your "squares" and play with the numbers:

i.e. 3 and 5

5x5= 25 3x3= 9

25-9 = 16 which is divisible by 4

i.e. 7 and 5

7x7= 49 5x5=25

49-25= 24 which is divisible by 4

Answer 3

enable the weird numbers be 2r+a million and 2s+a million, the position r and s are valuable entire numbers (2r-a million)^2 - (2s-a million)^2 = 4r^2 - 4r + a million - 4s^2 + 4s - a million = 4(r^2 - s^2) - 4(r-s) = 4(r-s)(r+s) - 4(r-s) = 4(r-s)(r+s-a million) all of us recognize that the adaptation of the sq. of two unusual numbers will be a particular of four from above. If r and s are both unusual or both even, (r-s) turns into even, i.e. a particular of two. If r and s are such that one is even and the different unusual, then (r+s-a million) is continually even, i.e. a particular of two. hence, (2r-a million)^2 - (2s-a million)^2 = 4(r-s)(r+s-a million) is continually a particular of 8.