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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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Solution Answer Comment 20 of 20 |
All odd numbers are of the form 4x+1 or 4x+3. We have two cases.

(4x+1)^2=16x^2+8x+1=8(2x^2+x)+1
n=2x^2+x

(4x+3)^2=16x^2+24x+9=8(2x^2+3x+1)+1
n=2x^2+3x+1

The pattern for n is triangular numbers.


  Posted by Math Man on 2013-06-10 16:51:54
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