+260 Prove algebraically that the square of any odd number is always 1 more than a multiple of 8. Let n stand for any integer in your working.
Thank you!
+129852 May be better ways of doing this....but.....here's my attempt
Any odd can be written as (2n + 1).where n is an integer...and a multiple of 8 can be written as 8m where m is an integer.....so we are trying to show that
(2n + 1)^2 = 8m + 1 subtract 1 from both sides
(2n + 1)^2 - 1 = 8m expand the left side
4n^2 + 4n + 1 - 1 = 8m
4n^2 + 4n = 8m factor
4n ( n + 1) = 8m
Show that the left side is true for n = 1
4(1) (1 + 1) = 4*2 = 8 which is a multiple of 8
Assume that it is true for n = k ....that is
4k (k + 1) = 8m
Prove it is true for n = k + 1
4(k + 1) [ (k + 1) + 1 ]
4 ( k + 1) [ k + 2 ]
4(k + 1) * k + 4(k+ 1)* 2
4k ( k + 1) + 8 (k + 1) [ 4k (k + 1) = 8m is true by assumption ]
8m + 8 ( k + 1)
8 ( m + k + 1)
Let (m + k + 1) = q
So
8q is a multiple of 8
