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You should do a series [heh] about being a "mathemagician".

I also found it very appropriate that I saw an ad for "WoW Gold" next to this comic.

I would use the following proof for Fermat's last theorem:

1) (...) No non-zero solution exists for n=3

2) The case of n>3 is left as an exercise for the reader.

It sure makes the proof a lot shorter. But rather harder to get published...

This is a good, but old joke: see the second one in the first row here.

Gasp! Magic! You must work in the goverment's budget office!

LMAO, I drew tiny "math fairies" on a calc test once to show how I reached an answer. For some reason, the professor marked it wrong...

I wrote this in an exam once. I could remember the first and last line of the proof, so the proof went something like:

line 1 of proof

= line 2 of proof

therefore line 3 of proof using Thm 2.3

...

magic happens!

...

= answer

Another time I ended my answer to a question with "I have a solution to this problem, but this exam is too short to explain it".

You should check out this book:

http://books.google.com/books?id=61BO2UY9QN4C&pg=PA65&lpg=PA65&dq=gamelin+conjure+up&source=bl&ots=5bpRixFd-o&sig=7aGXq7J4qdsf2cmr9PfV1qlNIiI&hl=en&ei=sgP3TLHPF8H68Aadkti9BQ&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBgQ6AEwAA#v=onepage&q&f=false

Close up if you miss it: http://i50.photobucket.com/albums/f348/XavvaX/Magic.png?t=1291256216

This proof should be interesting:

Prove that the nth root of 2 is irrational.

Proof by contradiction.

If it was rational, then (a/b)^n is equal to 2

a^n/b^n = 2

a^n = 2b^n

a^n = b^n + b^n

Fermat's last theorem proves this wrong.

Therefore, the nth root of 2 is irrational.