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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 Alternate Approach | Comment 4 of 20 |
X² - 1 = (X + 1)(X - 1)

If X is odd (X = 2a + 1), then (X + 1) and (X - 1) are even

X + 1 = 2a +2
X - 1 = 2a

(X + 1)(X - 1) = (2a + 2)(2a) = 4a² + 4a = 4a(a + 1). Since either a or (a + 1) is even, a(a + 1) is even and 4a(a + 1) is divisible by 8


To show the triangle number connection, it still can be shown that 4a² + 4a = 8*T(a)
  Posted by TomM on 2002-10-06 17:52:23
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