# 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?

• Solutions of $A^k + B^k = C^k$ in $n \times n$ integral matrices, American Mathematical Monthly, 75, 1968, 759-760. – Ethan Bolker Mar 26 '16 at 1:39
• @MXYMXY: the set of integers modulo $p$, or the set of congruence classes of same... It might be read "the integers modulo the ideal $p$ times the integers", or something like that – abiessu Mar 26 '16 at 2:02
• – Watson Mar 26 '16 at 9:42
• in the $p$-adic units Fermat fails for about 1/6-th of the primes. Also see this question here – ArtW Jun 20 '16 at 20:39
• For any $x,y,n$, it fails in the ring of integers of ${\bf Q}(\root n\of{x^n+y^n})$. Whether that ring is ever a UFD for $n\ge3$, I do not know. – Gerry Myerson May 13 at 5:26

$$(18+17\sqrt2)^3+(18-17\sqrt2)^3=42^3$$ 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 $${\bf Q}(\sqrt{-3})$$)

• Also for ${\bf Q}(\sqrt2)$ we have $(707472 + 276119\sqrt2)^3 + (707472 - 276119\sqrt2)^3 = 1106700^3$ – Josh B. May 13 at 16:02
• I'm surprised that $(1+\sqrt{-2})^3+(\sqrt{-2})^3=(1-\sqrt{-2})^3$ hasn't made it in here yet. – Oscar Lanzi May 13 at 23:11

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 comment by @ArtW on the original question suggests wholesale $p$-adic Fermat violations, and links to a discussion thereof. – Gerry Myerson May 13 at 22:23
• I saw that, and looked at the earlier question. Turns out there are solutions even for "bad" prime bases. For instance, the method above applies to $3^2+4^2=5^2$ in $59$-adics, where $3$ and $5$ both turn out to have square roots. – Oscar Lanzi May 13 at 23:21