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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!

smileysmileysmiley

 Feb 16, 2019
 #1
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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

 

 

 

cool cool cool

 Feb 16, 2019

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