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Square of an Odd (Posted on 2002-10-06) Difficulty: 2 of 5
Take any odd number and square it. It will invariably be a multiple of 8 plus 1. So (odd)^2=8n+1 where n is an integer. Show why this is always so. Also show what the pattern for n is.

See The Solution Submitted by martyn    
Rating: 3.1333 (15 votes)

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Some Thoughts hmm.. | Comment 15 of 20 |

i saw the solution..it looked over complicated...haha =0)

odd number = (2k + 1)... k belonging to Integer..

(2k+ 1)^2 = 4k^2 + 4k + 1
so...

4k^2 + 4k = 8n
k^2 + k = n
k(k+1) = 2n

the rest was the same...k's either even or odd..
this has prolly been posted already..but..*shrug* i'm too lazy to read thru all of the posts ^^ haha

ignore this if you wanna.


  Posted by Amon on 2005-05-04 02:46:33
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