2010-04-01

Math Proof: An attempted answer

I would like to thank all the people who answered my post on wanting a proof for the problem (thank you also to the people who hunted me down in IRC and pointed out what I needed to do also.) From all that help here is my attempt at a 'proof'.


Given:

In the series of non-negative integers, every number is divible
by one of the following: 3m, 3m+1, or 3m+2.

The series n^2 would thus be looked at as

(3m)^2 = 9m^2
(3m+1)^2 = 9m^2+6m+1
(3m+2)^2 = 9m^2+12m+4


9m^2 is always divisible by 3 as any number multiplied by 3 is
divisible by 3.
9m^2+6m and 9m^2+12 mis always divisible by 3 because a) but parts
of the sum are divisible by 3, and adding two numbers divisible
by 3 leaves you a number still divisible by 3.].. so we can simplify
this a bit

(3m)^2 = 9m^2 = 3j
(3m+1)^2 = 9m^2+6m+1 = 3j+3k+1 = 3(j+k)+1 = 3l+1 (or not divisible by 3)
(3m+2)^2 = 9m^2+12m+4 = 3j+3k+4 = 3(j+k)+4 = 3l+4
= 3l+3+1 = 3(l+1)+1 => 3n+1 (or not divisible by 3)

Thus all numbers squared are either divisible by 3 or 3x+1 (where x is
some number). And from that we can show that n^2+1 are never divisible
by 3 because they are either 3x+1 or 3x+2.



I would probably get an F from my favorite math professor Alan Sharples but it did make me feel a lot better when I got it out. Thanks for the help.. I can sleep easier tonight (well unless I get caught up in why Boron absorbs neutrons so well)
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