Fermat's Last Theorem near misses?

Solution 1:

The following shows that for $n=3$ the ratio can be as close to $1$ as we like. For any positive integer $m$ let:

$a = 3m^3 + 3m^2 + 2m + 1$

$b = 3m^3 + 3m^2 + 2m$

$c = 3m^2 + 2m + 1$

$d = m$

Then it is readily shown that $a^3 = b^3 + c^3 + d^3$. If we focus on the 'wrong solution' $a^3 = b^3 + c^3$, the ratio of LHS to RHS is:

$$\frac{a^3}{b^3+c^3}=\frac{a^3}{a^3-d^3}=\frac{1}{1-(d/a)^3}$$

Since $a$ increases much faster than $d$ as $m$ is increased, we can make the ratio as close to $1$ as we like by choosing a sufficiently large value of $m$. With $m=10$, for example, we have:

$$\frac{a^3}{b^3+c^3} = \frac{3321^3}{3320^3+321^3} = \frac{36627445161}{36594368000+33076161}=1.0000000273$$

Solution 2:

Noam Elkies gave many examples, as the output of a very fast algorithm for finding almost-solutions to arbitrary 3-variable homogeneous equations, in

http://arxiv.org/abs/math/0005139

The table on page 15 lists 37 examples for Fermat's Last Theorem with exponent between $4$ and $20$. Elsewhere in the article there are near-solutions of $x^\pi + y^\pi = z^\pi$ and many other interesting things.

Solution 3:

There are "almost" solutions of $a^3+b^3=c^3.$ If $3$ does not divide $a+b,$ and $a+b$ is factored out of $a^3+b^3,$ you get the equation $a^2-ab+b^2=T^3$ and there are solutions of this equation. What's interesting is that Furtwangler's and Vandiver's theorems (from the classical approach to FLT) are still applicable.