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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 Puzzle solution | Comment 17 of 20 |

Since the integer is odd let it be equal to (2p+1, say)
Then,
(2p+1)^2
= 8*n + 1, where n = p(p+1)/2

If p is odd, then p+1 is even, and so, n = p(p+1)/2 is an integer.
If p is even then clearly n must be an integer.

Thus n is always an integer irrespective of whether p is even or odd.

Edited on April 20, 2007, 12:16 pm
  Posted by K Sengupta on 2007-04-20 12:16:21

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