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)

## 2 comments:

Dude, typo:

"divible"

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

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.

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