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Did Homer Simpson Actually Solve Fermat's Last Theorem? Take A Look

Tuesday, May 13, 2014 - 12:53 PM



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Simon Singh is the author of The Simpsons and Their Mathematical SecretsBrady Haran's videos cover a great many subjects, mostly mathematical, presented by all kinds of wonderful scholars. You can find them collected here.





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Comments [9]

Hugh O Neill

He would have given a proof of the theory, showing it to be incorrect. Proofing and proving are not the same thing

May. 23 2014 07:20 AM
John Mitchell from Portland, OR

If you're curious by how much homer's solution actually missed by, a computer algebra system (or wolfram alpha online) gives the left hand side of the equation is:

3987^12 +4365^12 = 63976656349698612616236230953154487896987106

whereas the left hand side is

4472^12 = 63976656348486725806862358322168575784124416

which are the same up to the first 10 significant figures, so a calculator using scientific notation shows them as identical. However they actually differ by


which is a "near miss" of a million billion billion billion (!).

May. 21 2014 02:00 AM
j0equ1nn from New York

Yeah, to beat a dead horse, it's kind of like this.

Suppose there's an enormous bag of marbles, and suppose I conjecture that there are no red marbles in the bag. Then I reach into the bag, select a single marble, and show you that it's blue. You would not (should not) be convinced that I have proven my conjecture. I would need to verify that this would happen no matter which marble I drew.

Now suppose there are infinitely many marbles in the bag. Then no matter how many non-red ones I pulled out, one by one, we could never be certain about my conjecture. I would need to find some other, more categorical method of establishing that none of these marbles are red.

If the equation homer had written were true, it would, in this analogy, be the equivalent of someone reaching into the bag and pulling out red marble. This would certainly establish that I was wrong: case closed.

Fermat's theorem basically says that an equation of the form Homer has written cannot be true. But there are infinitely many equations of the form Fermat specified. One can check them one at a time to make sure they're false, but this would never settle the matter. If, on the other hand, one of these equations were found to be true, well then Fermat would have been wrong.

If you want to know about the actual math, don't be scared, look it up! It is not the kind of thing you need a PhD, or even more than a high school education, to understand the premise of. Number theory is kind of like that: simple questions, insanely complicated answers.

May. 17 2014 08:59 PM
Paul Hughes

Abbot and Costello — 7 X 13 = 28

May. 16 2014 12:42 PM
Jorge from Ireland

Actually, proving and disproving are very much the same thing. Disproving a statement means to prove its negation. What is usually the case, is that many a time a conjecture in math is written as "for all x, something about x". Its negation is only, "there exists an x so that something does not happen". Proving universality is certainly difference to existence.

May. 16 2014 01:43 AM
Caitlin from Los Angeles CA

DOCTOR Who (not Dr. Who).

<3 a whovian

May. 15 2014 04:20 PM

There is a HUUUUuuge difference between proving something & dis-proving it in math.

To prove Fermat's theorem you have to prove a^n + b^n =/= c^n for ALL values of n>2. People have been able to prove it for various specific values of n for quite some time. But the difficulty was to show that for ANY/ALL values of n.

To disprove on the other hand just requires 1 example. (like what Homer showed if it were true)

May. 15 2014 02:32 AM

Why would he only be able to disprove it? This is Mathematics, not science, and while scientists cannot prove things, Mathematicians can.

May. 14 2014 05:26 PM

IF it were true Homer would have "dis-proved" Fermat's theorem,
not proved it.

May. 14 2014 01:57 PM

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