Prove: if \"a is an odd integer\" then \" 8|(a^2-1)\" Solution If a is an odd integer, then a+1 is even and therefore b = (a+1)/2 is an integer. b = (a+1) / 2 implies a+1 = 2b or a = 2b-1 Let us calculate a^2-1: a^2 - 1 = [(2b-1)^2] - 1 and using identity (x-y)^2 = x^2 -2xy + y^2 on the (2b-1) term: a^2 - 1 = [4b^2 - 4b + 1] - 1 So: a^2 - 1 = 4b^2 - 4b + 1 - 1 a^2 - 1 = 4b^2 - 4b a^2 - 1 = 4b (b - 1) Now if b is even, and greater than zero, b can also be written as 2c where c = b/2 And if b is not even (odd) then b-1 is even (possibly b- 1=0, see note at the end) and can be written as 2c where c = (b-1)/2 and is greater than zero. (at least c=1). So we have 2 cases: (case 1:b even) then c = b/2 and: a^2 - 1 = 4(2c)(b - 1) = 8c(b-1) which is a multiple of 8; OR (case 2:b-1 even) then c = (b-1)/2 and: a^2 - 1 = 4(2c)(b) = 8cb which is a multiple of 8. Done but see note below: Note: IF b-1=0 this means a =1 and then, a^2-1 = 0, which is still 8 times zero ....

1. Prove: if "a is an odd integer"
then " 8|(a^2-1)"
Solution
If a is an odd integer, then a+1 is even and therefore b = (a+1)/2 is an integer. b =
(a+1) / 2 implies a+1 = 2b or a = 2b-1 Let us calculate a^2-1: a^2 - 1 = [(2b-1)^2] - 1 and using
identity (x-y)^2 = x^2 -2xy + y^2 on the (2b-1) term: a^2 - 1 = [4b^2 - 4b + 1] - 1 So: a^2 - 1 =
4b^2 - 4b + 1 - 1 a^2 - 1 = 4b^2 - 4b a^2 - 1 = 4b (b - 1) Now if b is even, and greater than zero,
b can also be written as 2c where c = b/2 And if b is not even (odd) then b-1 is even (possibly b-
1=0, see note at the end) and can be written as 2c where c = (b-1)/2 and is greater than zero. (at
least c=1). So we have 2 cases: (case 1:b even) then c = b/2 and: a^2 - 1 = 4(2c)(b - 1) = 8c(b-1)
which is a multiple of 8; OR (case 2:b-1 even) then c = (b-1)/2 and: a^2 - 1 = 4(2c)(b) = 8cb
which is a multiple of 8. Done but see note below: Note: IF b-1=0 this means a =1 and then,
a^2-1 = 0, which is still 8 times zero ...