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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 Heh! | Comment 1 of 20
1² = 1 = 8(0) + 1
3² = 9 = 8(1) + 1
5² = 25 = 8(3) + 1
7² = 49 = 8(6) + 1
9² = 81 = 8(10) + 1
11² = 121 = 8(15) +1

"n" always seems to be a triangle number: T(x) = 1 + 2 + 3 +...+ x = [x(x + 1)]/2

Explanation:

Any odd number is of the form X = 2x + 1
X² = (2x + 1)²
X² = 4x² + 4x + 1
but 4x² + 4x = 8[(x² + x)/2] = 8 * T(x)

So X² = [8 * T(x)] + 1
  Posted by TomM on 2002-10-06 13:25:45
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