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

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Fermat’s

for

Yet in an episode of The Simpson’s it was noted that , apparently contradicting Fermat’s last theorem.

Of course this is not a contradiction because is not actually equal to .

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

Well,

while

So, starting from the left we see that the digit in is a 9, while for 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?”

x

The left most digits of these large numbers represent very large quantities.

x

Maybe one way to compare these two numbers is to look at their ratio .

x

If we have a calculator that’s accurate to 10 decimal places we get the result

x

x

So to 10 decimal places the ratio is 1.

x

It’s not until we get a more accurate calculator that we see, in fact,

x

x

Another way to think about the relationship between and is to take the twelfth root of .

x

If we really had equality between and then would be .

x

If we have a calculator that’s accurate to 12 digits, we get

x

It’s only with a more accurate calculator that we see

x

This raises a sort of interesting question:

x

Given a “small” potential error and an integer , what solutions are there to

x

x

for positive integers ?

38305^3 + 51762^3 = 57978^3

49193^3 + 50920^3 = 63086^3

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

1 | Huenyk

January 26, 2013 at 9:41 pm

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?