Near misses in Fermat’s last theorem

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

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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?”

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

$a^n+b^n=(c+\epsilon)^n$

for positive integers $a, b, c$ ?

Republic of Mathematics blog