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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!

qualitystreet Feb 16, 2019

#1**+3 **

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

CPhill Feb 16, 2019