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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 Answer to first question | Comment 5 of 20 |
if x is an integer

x(x+1) = an even integer,we'll call it 2n where n is an integer)

So, x(x+1) = 2n
x^2+x = 2n
4x^2 + 4x = 8n
4x^2 + 4x + 1 = 8n + 1
(2x + 1)^2 = 8n+1

but 2x + 1 is odd so (2x+1)^2 = (odd)^2 = 8n + 1

  Posted by Dulanjana on 2002-10-07 14:46:57
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