Republic of Mathematics blog

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?


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


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

3 Responses to "Near misses in Fermat’s last theorem"

You can find thousands of near-misses by solving the following equation:
A^n+B^n = C^n + !
or A^n+B^n = C^n – !
The difference is always !. I tested these up to n = 1000 and these still hold.
Surely based on asymptotic argument, these are counter-examples to FLT?

38305^3 + 51762^3 = 57978^3
49193^3 + 50920^3 = 63086^3

I think I found a set of formulas which work to find the series of best “near solutions” to Fermat’s Last Theorem out to infinity. I can’t be sure … just have an ordinary PC and Lotus spreadsheet which is accurate
only to 18 digits. Anyone who would like to see the formulas can email me … I’ll attach my final spreadsheet, which does work for 55 equations in a row … I’m sure of that. HH

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