However, it's actually not too hard to disprove 3987^12 + 4365^12 = 4472^12 with just a calculator. The thing to do is concern yourself with the rightmost digits, not the leftmost.
The fake solution is actually a bit better than this article indicates, since checking only the rightmost digit -- working modulo 10 -- doesn't show the problem. 7^12 ends in 1, and 5^12 ends in 5. So the value of the left-hand side of the proposed equation ends in 6. And so does 2^12.
But checking the rightmost 2 digits -- working modulo 100 -- we can see that it doesn't work. 87^12 ends in 81, and 65^12 ends in 25. Their sum ends in 06. But 72^12 ends in 16. So the proposed equation is false.
-----
EDIT. For anyone who wonders how to do this.
Raising a number to the 3rd power, and then raising that to the 4th power, we get the 12th power of the original number. If we are working modulo 100, then we can throw out all but the rightmost 2 digits at any stage.
So, what are the rightmost 2 digits of 3987^12? Start by finding 87^3: 658503. We only care about the "03" at the end. 03^4 = 81. So 3987^12 ends with 81.
Similarly, for 4365^12: 65^3 = 274625. 25^4 = 390625. So 4365^12 ends with 25.
And, for 4472^12: 72^3 = 373248. 48^4 = 5308416. So 4472^12 ends with 16.
Another simple disproof: 3987 and 4365 are both divisible by 3 (and thus so is the left-hand side), but 4472 is not (and thus neither is the right-hand side, as 3 is prime).
Another easy one is looking at it modulo 4. The powers on the left-hand side are both odd perfect squares, which are thus 1 modulo 4. So their sum is 2 modulo 4. But the right-hand side is an even perfect square, which is 0 modulo 4.
EDIT. I guess my modulo-100 discussion is really just a convoluted way of coming up with the above fact.
Homer³ — the 3D sequence in the Halloween episode Treehouse of Horror VI, three years previous — had a similar ‘counterexample’: 1782¹²+1841¹²=1922¹². And also, 1+1=2, e^iπ=-1, and P=NP.
His real middle name is Samuel. He had to change his middle initial because there was already a member of the Writers Guild of America with the name "David S Cohen".
1. You mean "David Samuel(S.) Cohen", one of the writers for The Simpsons and Futurama, who decided to make his name "David X. Cohen" (not his legal name) because the Writer's Guild does not accept duplicate names?
2. Yes, people do believe he had a proof somewhere in his mind, considering the rest of his track record. Whether that proof was correct is definitely up for debate.
Well, according to a mathematician friend of mine, he definitely did not. He went to a lot of trouble later in his life to prove things that would have followed trivially if he had had such a proof. Besides, the result is interesting enough that he obviously would have published the proof if it existed. Most likely he thought he had a proof when he made the margin note, but later found a flaw. He would have felt no need to issue a retraction, since he had never formally claimed to have a proof in the first place.
Without having seen the proof that came to Fermat's mind, I don't think it's anyone's absolute opinion on the matter that "It's not up for debate" and that Fermat's proof being potentially true is "missing the fundamentals of science".
If you may re-read my previous post, here's the thing people disagree on : "Did Fermat write this having a correct proof of it on his mind or not?" I do not think anyone can say an absolute "yes" or "no" to this. Although it does point to "no", considering that the mathematics used to prove FLT are far beyond Fermat's time.
Given the eventual proof of Fermat's last theorem there's a solid argument that Fermat hadn't successfully proved it (and certainly couldn't have in the way it was eventually done), but it's also far from clear whether he actually believed he had.
Another interesting easter egg is the Futurama Theorem (http://theinfosphere.org/Futurama_theorem). The writers for some of these shows are brilliant people who happen to write for cartoons.
As an aficionado of both math and The Simpsons myself I quite liked the author's book "The Simpsons and their Mathematical Secrets". One thing to note before reading: it's targeted at an audience with a passing familiarity with The Simpsons, so an unfortunate percentage of it is a description of episodes you probably already know by heart. But it's a good read, anyway, so check it out at your local library.
20 comments
[ 5.5 ms ] story [ 54.0 ms ] threadHowever, it's actually not too hard to disprove 3987^12 + 4365^12 = 4472^12 with just a calculator. The thing to do is concern yourself with the rightmost digits, not the leftmost.
The fake solution is actually a bit better than this article indicates, since checking only the rightmost digit -- working modulo 10 -- doesn't show the problem. 7^12 ends in 1, and 5^12 ends in 5. So the value of the left-hand side of the proposed equation ends in 6. And so does 2^12.
But checking the rightmost 2 digits -- working modulo 100 -- we can see that it doesn't work. 87^12 ends in 81, and 65^12 ends in 25. Their sum ends in 06. But 72^12 ends in 16. So the proposed equation is false.
-----
EDIT. For anyone who wonders how to do this.
Raising a number to the 3rd power, and then raising that to the 4th power, we get the 12th power of the original number. If we are working modulo 100, then we can throw out all but the rightmost 2 digits at any stage.
So, what are the rightmost 2 digits of 3987^12? Start by finding 87^3: 658503. We only care about the "03" at the end. 03^4 = 81. So 3987^12 ends with 81.
Similarly, for 4365^12: 65^3 = 274625. 25^4 = 390625. So 4365^12 ends with 25.
And, for 4472^12: 72^3 = 373248. 48^4 = 5308416. So 4472^12 ends with 16.
81 + 25 = 106, which does not end with 16.
Another easy one is looking at it modulo 4. The powers on the left-hand side are both odd perfect squares, which are thus 1 modulo 4. So their sum is 2 modulo 4. But the right-hand side is an even perfect square, which is 0 modulo 4.
EDIT. I guess my modulo-100 discussion is really just a convoluted way of coming up with the above fact.
I'm so old I remember when The Simpsons was good.
2. Does anyone believe Fermat had a proof?
This article just smells sloppy.
http://en.wikipedia.org/wiki/David_X._Cohen#Name_change
2. Yes, people do believe he had a proof somewhere in his mind, considering the rest of his track record. Whether that proof was correct is definitely up for debate.
No, no it's not.
You're not really understanding mathematics, which is fair enough, but also the fundamentals of science and human nature with a statement like that.
Without having seen the proof that came to Fermat's mind, I don't think it's anyone's absolute opinion on the matter that "It's not up for debate" and that Fermat's proof being potentially true is "missing the fundamentals of science".
If you may re-read my previous post, here's the thing people disagree on : "Did Fermat write this having a correct proof of it on his mind or not?" I do not think anyone can say an absolute "yes" or "no" to this. Although it does point to "no", considering that the mathematics used to prove FLT are far beyond Fermat's time.
Given the eventual proof of Fermat's last theorem there's a solid argument that Fermat hadn't successfully proved it (and certainly couldn't have in the way it was eventually done), but it's also far from clear whether he actually believed he had.
"The Simpsons and their Mathematical Secrets"
http://www.simonsingh.net/Simpsons_Mathematics/book/