Prove A If m and n are integers and mn is even then mn is ev

Prove:

A) If m and n are integers and m+n is even, then m-n is even.

B) If n is an odd integer, then 8 divides n^2 - 1.

Solution

#(A)  

If m+n is even then it will be divisible by 2.

So, If m + n is divisible by 2 (i.e., an even integer), then m - n is also divisible by 2 because: m+n = m-n + 2n, which equals 2k + 2n = 2(k+n).

#(B)

An odd integer n is either a 4k +1 or a 4k+3

so, (4k + 1)² -1 = 16k² +8k + 1 - 1

                         = 16k² +8k

                          = 8(2k + k ) which is a multiple of 8.

and (4k + 3)² -1 = 16k² + 24k + 9 - 1

                  = 16k² + 24k + 8

                = 8( 2k² + 3k + 1 )

which is a multiple of 8.    [proved]