Where Fermat's last theorem fails It's fairly well known that Fermat's last theorem fails in $\mathbb{Z}/p\mathbb{Z}$. Schur discovered this while he was trying to prove the conjecture on $\mathbb{N}$, and the proof is an application of one of his results in Ramsey theory, now known as Schur's theorem. 
I'm wondering whether there are any other places (let's say, unique factorisation domains) where the statement is known to be false?
 A: You can also blow FLT out of the water in $p$-adics.  Consider the ordinary Pythagorean triple
$17^2+144^2=145^2$
Render these arguments in $2$-adics:  $17$ and $145$ are each one greater than a multiple of $8$, thus squares of other $2$-adic integers which I shall call $\pm m$ and $\pm n$ respectively (an additive inverse pair of choices for each).  And of course $144$ is the square of $\pm 12$.  So then we have eight $2$-adic equations (four of them "linearly independent") of the form
$(\pm m)^{\color{blue}{4}}+(\pm 12)^{\color{blue}{4}}=(\pm n)^{\color{blue}{4}}$
A: $$(18+17\sqrt2)^3+(18-17\sqrt2)^3=42^3$$ so Fermat fails for $n=3$ in the UFD ${\bf Q}(\sqrt2)$. $$(1+\sqrt{-7})^4+(1-\sqrt{-7})^4=2^4$$ so Fermat fails for $n=4$ in the UFD ${\bf Q}(\sqrt{-7})$.
Looking at a couple more of the imaginary quadratic UFDs:
$(1+\sqrt{-2})^{\color{blue}{3}}+(\sqrt{-2})^{\color{blue}{3}}=(1-\sqrt{-2})^{\color{blue}{3}}$ (failure in ${\bf Q}(\sqrt{-2})$)
$(1+\sqrt{-3})^{\color{blue}{6n+1}}+(1-\sqrt{-3})^{\color{blue}{6n+1}}=2^{\color{blue}{6n+1}}$ (any whole number $n$ at all; failure in ${\bf Q}(\sqrt{-3})$)
