Republic of Mathematics blog

Sanjoy Mahajan – Rote Learning Fragments the World

Posted by: Gary Ernest Davis on: March 2, 2011

I had the pleasure of interacting with Sanjoy many years ago at Rutgers, stuff New Brunswick, treat when he was doing research on Louis Benezet.

Sanjoy is a fabulous problem solver.

Near misses in Fermat’s last theorem

Posted by: Gary Ernest Davis on: March 2, 2011

Fermat’s last theorem, proved by Andrew Wiles, states that there are no positive whole number solutions for a, b, c to the equality

a^n+b^n=c^n for n\geq 3

Yet in an episode of The Simpson’s it was noted that 3987^{12} + 4365^{12} = 4472^{12}, apparently contradicting Fermat’s last theorem.

Of course this is not a contradiction because 3987^{12} + 4365^{12} is not actually equal to 4472^{12}.

So why did The Simpson’s episode say these two numbers were equal?

Well,

3987^{12} + 4365^{12} =63976656349698612616236230953154487896987106

while

4472^{12} = 63976656348486725806862358322168575784124416

 

So, starting from the left we see that the 11^{th} digit in 3987^{12} + 4365^{12} is a 9, while for 4472^{12} it is an 8.

David Radcliffe (@daveinstpaul ) tweeted:

“What is the best way to measure the “closeness” of a near-solution to a^p+b^p=c^p? E.g. how “close” is 3987^12 + 4365^12 = 4472^12 really?”
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The left most digits of these large numbers represent very large quantities.
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Maybe one way to compare these two numbers is to look at their ratio \frac{3987^{12} + 4365^{12}}{4472^{12}}.
x
If we have a calculator that’s accurate to 10 decimal places we get the result
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\frac{3987^{12} + 4365^{12}}{4472^{12}} =1.0000000000
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So to 10 decimal places the ratio \frac{3987^{12} + 4365^{12}}{4472^{12}} is 1.
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It’s not until we get a more accurate calculator that we see, in fact,
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\frac{3987^{12} + 4365^{12}}{4472^{12}} \approx 1.0000000000189426406
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Another way to think about the relationship between 3987^{12} + 4365^{12} and 4472^{12} is to take the twelfth root of 3987^{12} + 4365^{12}.
x
If we really had equality between 3987^{12} + 4365^{12} and 4472^{12} then (3987^{12} + 4365^{12})^{1/12} would be 4472.
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If we have a calculator that’s accurate to 12 digits, we get (3987^{12} + 4365^{12})^{1/12}= 4472.00000000
x
It’s only with a more accurate calculator that we see (3987^{12} + 4365^{12})^{1/12}\approx 4472.0000000070592907
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This raises a sort of interesting question:
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Given a “small” potential error \epsilon and an integer n\geq 3, what solutions are there to
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a^n+b^n=(c+\epsilon)^n
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for positive integers a, b, c ?