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 Go for short! (Posted on 2005-04-19)
Looking at the "Square of an Odd" puzzle that asks to prove that the square of an odd number is always 1 more than a multiple of 8, a professor gave this four parts proof: "All odd numbers are of the form 8K+1, 8K+3, 8K+5 or 8K+7. Squaring these numbers produces 8M+1, 8M+9, 8M+25 or 8M+49, which are all of the form 8N+1. QED"

Another professor came by, and gave a single line proof. Can you manage it?

Note: no one who answered the original problem produced either the four parts solution, or the single line one.

 See The Solution Submitted by e.g. Rating: 2.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 Another way | Comment 2 of 8 |
E.G.'s professor wrote odd numbers as 8K+something; Bractal wrote 2K+1. I chose writing them as 4K+T, where T is either +1 or -1. As T²=1, then (4K+T)²= 16K²+8KT+1= 8(2K²+KT) + 1.

Edited on April 19, 2005, 12:46 pm
 Posted by Federico Kereki on 2005-04-19 12:43:14

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