I've recently seen a video of Numberphille channel on Youtube about Fermat's Last Theorem. It talks about how there is a given "solution" for the Fermat's Last Theorem for $n>2$ in the animated series The Simpsons.

Thanks to Andrew Wiles, we all know that's impossible. The host tells that the solution that appears in the episode is actually a near miss.

There are two near-miss solutions and they are:

$$3987^{12} + 4365^{12} \stackrel{?}{=} 4472^{12}$$

$$1782^{12} + 1841^{12} \stackrel{?}{=} 1922^{12}$$

Anyone who knows modular arithmetic can check that those solutions are wrong, even without a calculator. You can check that $87^{12} \equiv 81 \pmod{100}$ and $65^{12} \equiv 25 \pmod {100}$, while $72^{12} \equiv 16 \pmod {100}$. So we have: $$ 81 + 25 \stackrel{?}{\equiv} 16 \pmod {100} $$ $$ 106 \stackrel{?}{\equiv} 16 \pmod {100} $$ which is obviously wrong.

For the second example it's even easier. We know that LHS is an addition of an even and odd number, and the RHS is even number, which is impossible, because we know that the addition of an even and an odd number will provide an odd number.

But what's made me most interested in this is the following. Using a calculator I expanded the equations and I get:

$$3987^{12} + 4365^{12} \stackrel{?}{=} 4472^{12}$$

$$63976656349698612616236230953154487896987106 \stackrel{?}{=} 63976656348486725806862358322168575784124416$$

$$1211886809373872630985912112862690 \stackrel{?}{=} 0$$

And you'll immediately conclude that those numbers aren't even close, their difference is a 33-digit number. But bearing in mind that we are working with really, really big numbers, it's probably better to take relative difference. So we really want to find the ratio of LHS and RHS: $$ \frac{63976656349698612616236230953154487896987106}{63976656348486725806862358322168575784124416} \approx 1.00000000002 $$ Somehow that's impressive, but if take a look into the second example the things start to get more interesting:

$$ 1782^{12} + 1841^{12} \stackrel{?}{=} 1922^{12} $$ $$ 2541210258614589176288669958142428526657 \stackrel{?}{=} 2541210259314801410819278649643651567616 $$

As we can see the first 9 digits are exactly the same and the apsolute difference is: $700212234530608691501223040959$. But if we take a relative difference or ratio we'll get: $$ \frac{2541210258614589176288669958142428526657}{2541210259314801410819278649643651567616} \approx 0.9999999997244572 $$ And this is pretty amazing, because if we make comparison using smaller number this is the same as comparing $10\, 000\, 000\, 000$ and $10\, 000\, 000\, 001$

Probably there are many more, probably infinite amount of such "close" examples, but are there any known examples? Is there any list of them?

And as user17762 commented it would be nice to find a bound $\phi(n) = \inf\{\left| x^n + y^n - z^n\right| : x,y,z \in \mathbb{Z}\}$, although I would be more interested in finding the ratio bound, such that the ratio is closer to $1$

Also as user17762 pointed Taxicab can be used to provide a really close examples for $n=3$, but what about other values for $n$?