Compatibility of probabilistic axioms and arithmetic - Fermat's Last Theorem

Question

From this answer I followed a link to this description of a proof which cites from this original source.

A "proof" of Fermat's Last Theorem is presented using manipulation of the equation according to theory of probabilities and De Morgan's laws, and it is difficult for me to spot where the reasoning goes off the rails—though it's obvious to me that it must have done so somewhere to reach the false conclusion.

Where is the flaw in this "proof's" logic?

Answer 1

The cleverness in arranging the false proof in the second link is that of blithely using $+$ to disguise the fact that you are really talking about $\cup$ and $\cap$ (union and intersection). Working with real numbers it is easy to go from $AB=0$ to "either $A=0$ or $B=0$."

Working with events or probabilities that reasoning is not valid. It is indeed very easy to find events $M$ and $N$ such that $P(M \cap N) = 0$ yet neither $P(M)$ nor $P(M)$ is zero.

For example, choose an integer $i$ uniformly randomly on $[1,10]$. Let event $M$ be "$4<i<9$" so $P(M) = \frac25$. Let event $N$ be "$i$ is a perfect square " so $P(N)=\frac{3}{10}$. Here, $P(M\cap N) = 0$.

I would also have to say that for two reasons it is prohibitively unlikely that Fermat had this proof in mind. The first is that its flaw is so easy to spot; Fermat was neither careless nor dull. The second is that it would easily have fit in the margin; much longer proofs were provided in that way.

Personally, I believe that Fermat's flawed proof was the one that would work if all complex integer fields were unique factorization domains; that is the sort of mistake (not anticipating a subtle and surprising future discovery) that a first-rate mathematician is vulnerable to.

Answer 2

The proof is complete nonsense. The first problem (or at least one of the first problems; there's a lot broken there) is that, in general, \begin{align*} P(\text{$X$ and $Y$}) &\not = P(X)P(Y); \\ P(\text{neither $X$ nor $Y$}) &\not = P(\text{not $X$})P(\text{not $Y$}) \end{align*} And it should be obvious that that approach doesn't work; FLT does have many real solutions, just not in integers.

If the proof were trying (and, again, it's not) to show that the probability that a point $(x,y,z)∈\mathbb{Z}$ chosen at random (whatever that means for $\mathbb{Z}$) satisfies $x^n+y^n=z^n$ for $n>2$, then that wouldn't imply FLT either. We already know (e.g., from Falting's theorem) that there are only finitely many integer solutions; the trick is showing that there aren't any nontrivial ones at all.