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Did Homer just disprove Fermat's Last Theorem? O.o
It looks like general relativity he's doing there
I meant line 2 (it's actually a joke the math professors on the writing team of the Simpsons put in there: it doesn't disprove Fermat's Theorem, but most calculators at the time didn't have the accuracy to cumpute that directly, which is kind of the joke)
Oh well spotted, you're right. I didn't twig at all.
There's a book on the subject by Simon Singh, where he writes about all the math jokes in the Simpsons. There are a bunch. ;)
M(Ho)=π
(Ho)M=π
HoM=π
HoMEr
No, but you just discovered a hidden nerd-bait joke from the show's creators. One of the guys working on the show created some sort of program that could generate close-but-not-quite solutions. The one shown here isn't actually equal, but they are close enough that the difference won't show up without a more precise calculator, since both sides are roughly 6.397665635 x 1043.
398712 + 436512 = 63 976 656 349 698 612 616 236 230 953 154 487 896 987 106
447212 = 63 976 656 348 486 725 806 862 358 322 168 575 784 124 416
EDIT: I should have read the other comments. Looks like I'm late to the party.
That addition of exponents looks wrong.