## 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 divibleby 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+49m^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 issome number). And from that we can show that n^2+1 are never divisibleby 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)

L said...

Dude, typo:

"divible"

(I'm no longer a true geek, now I'm a crazy proof-reading historian, sigh).

Stephen Smoogen said...

A true geek would always spell check another geek's work.. and there are plenty of historian geeks. Watch a History of Britain by Simon Shama for a real geek look at Britain.