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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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Square of an Odd Number | Comment 6 of 20 |
let m = any integer
then N = 2m + 1 is an odd number
N^2 = (2m+1)^2
= 4m^2 + 4m + 1
= 4m(m+1) + 1
= 8(m(m+1)/2) + 1
However m! = m(m+1)/2 where m! is m factorial
so
N2 = 8m! + 1
and n=m!
  Posted by Douglas Johnson on 2002-10-07 15:47:22
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