## Wednesday, November 5, 2008

### Fermat's "margin" PROOF!

Fermat's theorem was supposedly cracked by an English mathematician a few years back using math SOOOO complicated that they couldn't begin to explain it on the NOVA show about it.

I spent quite a few 'odd hours' puzzling over why it would be 'SOOOO' complicated. I seemed to always get stuck when I would ask myself that one damning question, "Why does that solution mean that it only works for whole numbers? It never did!

My attempted solution was always along the lines of, if there is a whole number

Solution to x(cubed) + y(cubed) = z(cubed) all being whole numbers then there couldn't be a whole number solution to yx(squared) + y(cubed) = yz(squared), which there obviously IS!

Seriously now, there is no way in HELL that, if the second equation is 'true' that you could decrease the value of y to x and also increase the value of y to z!(THAT IS FERMAT'S PROOF that fits in the margin, that no-one til I, pboyfloyd figured it out!)

I think that there needs to be a visual aid in the real-life not complicated solution because of the difficulty defining a whole number.

Picture a room with a tile floor. The tiles are layed first one white one in the corner to represent the 1(square)(duh!).

Three black tiles(unit squares) 'surround' the white tile making a four tile square representing 2 squared(or 4).

Five white tiles are layed 'surrounding' the "2 squared" to represent "3 squared"

All this is a question of 'can you "see"?

The 'stumbling block' is ... do the sides need to follow the 'rules' for squares?

If you 'see' it, YES THEY DO!

Here's the proof!

Imagine that there IS a solution to x(cubed) +y(cubed) = z(cubed)

This equals x times x(squared) + y times y(squared) = z times z(squared)

But to rearrange this equation to a similar one which we KNOW is true :-

k times x(squared) + k times y(squared) = k times z(squared)

all we need to do is make k (any constant whole number) equal to x or y or z.

But if we do that then we see that we are ADDING to the left side of the equation and SUBTRACTING from the right side of the equation!

This happens no matter what 'power' above 2(above squaring).

How about the power of 2, does it work there?

3 + 4 = 7 (let k = 4)

k3 + k4 = k7

12 + 16 = 28

Subtract 3 from both sides...

9+16=25

This becomes essentially a 'trap' because we can see that if we multiply by k (again letting k be the same as 'y', that is 4 in this scenario, we get:-

k9 + k16 = k25 or..

36 + 64 = 100

forcing another solution of power 2!

Attempting to jam a solution for power 3 means subracting 9 to make the first term 3^3 from the left side of the equation but now we must ADD 25 to the right side of the equation.

Attempting solutions for powers higher still only results in EVEN MORE ludicrous disparity!

The visualization part comes in when looking at what 'happens' when the third dimension is added.

\$\$\$
\$\$\$
\$\$\$
+

####
####
####
####

(# and \$ Both being unit squares)

=

\$\$\$\$\$
####\$
####\$
####\$
####\$
MUST follow this 'look' whether we are dealing with just the 'squares' or with ONE FACE of the cubes!

The 'tiles' earlier looked like this(BTW)

#\$#\$#\$#\$#\$#\$#\$#\$#\$#\$#
\$\$#\$#\$#\$#\$#\$#\$#\$#\$#\$#
###\$#\$#\$#\$#\$#\$#\$#\$#\$#
\$\$\$\$#\$#\$#\$#\$#\$#\$#\$#\$#
#####\$#\$#\$#\$#\$#\$#\$#\$#
\$\$\$\$\$\$#\$#\$#\$#\$#\$#\$#\$#
#######\$#\$#\$#\$#\$#\$#\$#
\$\$\$\$\$\$\$\$#\$#\$#\$#\$#\$#\$#
#########\$#\$#\$#\$#\$#\$#
\$\$\$\$\$\$\$\$\$\$#\$#\$#\$#\$#\$#
###########\$#\$#\$#\$#\$#
\$\$\$\$\$\$\$\$\$\$\$\$#\$#\$#\$#\$#
#############\$#\$#\$#\$#
\$\$\$\$\$\$\$\$\$\$\$\$\$\$#\$#\$#\$#
###############\$#\$#\$#
\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$#\$#\$#
#################\$#\$#
\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$#\$#
###################\$#
\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$#
#####################
(and so on down as far as you need)

if any added layer is a complete square, if any 'L' (or backward L in this case) of black or white tiles) then x^2 +y^2 = z^2

There are also a huge amount of 'non-solutions'. i.e. 7^2 + 7 along + the corner 1 + 7 up whch is 8^2 but 15 is not itself a 'square'.(hey, almost tho!)

Then there is the 'double layer' solutions. 8^2 plus the 9th and the 10th layer = 10^2

AND the the ninth layer of tiles plus the tenth layer add to 36, a perfect square(6^2)!

So, the general expression of true(and the trivial non-) solutions are x^2 + n2x+n^2 = z^2!(where n is a whole number duh!)

Or, visually pick a square on the floor made up of black and white rows and colunms, add the next row/column or the next two rows/columns or the next 3 etc. etc.

Example x^11 + y^11 = z^11 cannot be 'reconciled' with the 'truth' of y*x^10 + y*y ^10 equaling y*z^10!

Saint Brian the Godless said...
This comment has been removed by the author.
Saint Brian the Godless said...

Well I wasn't sure if I should do a post on kaballah on my blog, but if you can post this arcanum whose target audience is like two hundred people nationally, I guess I'm safe. :-)

Wish I could add something pertinant, but it's all abyssinnian to me.

pboyfloyd said...

Hey Brian, you can write about anything at ALL on your blog.

I will read it, consider it and comment on it.

You know, unless it's about kaballah, of course! LMAO

Asylum Seeker said...

My thoughts? The mathz...they burnz me!!!1!! Anonymous said...

Your blog keeps getting better and better! Your older articles are not as good as newer ones you have a lot more creativity and originality now keep it up! Anonymous said...

So I'm guessing 1 and -1 etal with z then equaling zero is not the answer? Would be too simple.