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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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a simple solution | Comment 11 of 20 |

represent odd # as 2K+1

then odd^2-1 =(2K+1)(2K+1)-1

4k^2+4K =4K(K-1)

but since either K or K-1 is even a 2 can be pulled out from one of them leaving 4K(k-1)/8 as an integer

  Posted by red_sox_fan_032003 on 2004-03-03 17:56:18
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