12
$\begingroup$

I came across this one of many claimed elementary proofs of FLT. It looked credible, and I felt surprised seeing that it wasn't drawing much attention from anyone. I investigated and I ended up finding this argument ruling out any chances for this kind of "proofs" to be correct. Now, my question shall be organized in three steps:

  • I have a very basic understanding of the "trick". I get its underlying logic, but unluckily I have a ridiculous knowledge of rings and fields, and in particular I know almost nothing of $p$-adic numbers. Can you confirm that, reasoning in analogy with the familiar number sets, I can safely assume that having a solution in $Q_p$ would imply a counterpart in $Z_p$? Is there a way to make it comprehensible how a solution made of $p$-adic integers would look like?

  • This is what I'm interested in most. Is it possible to understand, at least at an intuitive level, in lay-person's terms, what is the characteristic of the ring of familiar integers that makes it different from other factorial rings? In other words, what is (or are) the features typical only of our beloved usual numbers that make FLT hold for them? In further other words, what properties have been involved by the advanced mathematical tools used to prove FLT?

  • Finally, if we can spot such a characteristic, and it necessarily requires the use of instruments that Fermat didn't have, isn't this definitive evidence that Fermat could not have a proof? Why is this still sometimes questioned? Did he have any chance to perform something that did not fall under the disproof of the "trick", something that wouldn't apply to other rings like that of $p$-adic integers?

$\endgroup$
  • 3
    $\begingroup$ "having a solution in $Q_p$ would imply a counterpart in $Z_p$": yes, this is true, because one can "clear denominators" (in this case, multiply by an appropriate power of $p$) in $Q_p$ just as one can in $Q$ itself. $\endgroup$ – Greg Martin May 20 '17 at 0:06
  • 3
    $\begingroup$ Another property of $\mathbb Z$ that is not true in $\mathbb Z_p$ is the infinitude of primes. There is essentially on one prime in $\mathbb Z_p$. $\endgroup$ – Thomas Andrews May 20 '17 at 0:17
  • $\begingroup$ Oohh, so chances are that Krasner's trick isn't reliable? (I didn't find much googling it, I think that the book "Krasner versus Fermat" might be the reference) If this is the case, can I again accept the idea that an elementary proof of FLT might one day be found? $\endgroup$ – lesath82 May 20 '17 at 0:19
  • 5
    $\begingroup$ My motivation is that "layman" is gender-specific, while "layperson" is gender-neutral and thus more inclusive. $\endgroup$ – Greg Martin May 20 '17 at 1:21
  • 1
    $\begingroup$ Oh, and I heard that the entry in Bachet's edition of Diophantus concerning the equation $x^n + y^n = z^n$ was made by Fermat's wife. $\endgroup$ – franz lemmermeyer May 20 '17 at 5:51
8
$\begingroup$

I think the claim of that thread is blatantly overstated. For one thing, there are lots of properties that $\mathbb{Z}$ has, but $Z_p$ does not have. First and foremost, well ordering of the positive elements, which is heavily used as a key to solving many diophantine equations.

Now consider Fermat's elementary proof that $$x^4 + y^4 = z^4$$ has no solutions for $(x,y,z) \in \mathbb{Z}$ with $xyz \ne 0$.

I'm not sure whether or not there is a solution in $p$-adic integers, for some prime $p$, but if such solutions exist, it's an example showing that the existence of qualifying $p$-adic solutions doesn't imply the existence of qualifying integer solutions.

As I indicated in the comments, I suspect that most flawed attempts, at least the ones where the solver knows some Number Theory and is not obviously crazy, would include steps for which there is no analogue in $Z_p$, so the $Z_p$ criterion would be useless for invalidating those attempts.

For a given proposed proof, a more common way of quickly demonstrating that there must be a mistake$\,-\,$without actually pinpointing the error, is to observe that the argument would still work for the equation $x^2 + y^2 = z^2$, or, alternatively, to apply the argument line by line for the equation $x^3+y^3=z^3$, and see if the proof at least works for that case.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I don't think that argument is overstated, it is just hard to understand if you don't read carefully. Basically, he's just saying, "If the argument would work in $\mathbb Z_p$, it doesn't work." It doesn't mean (or say) there is no elementary proof, except for a very specific (and spelled out) definition of "elementary." $\endgroup$ – Thomas Andrews May 20 '17 at 0:26
  • 1
    $\begingroup$ I read it as: if the argument doesn't work in $Z_p$, then it doesn't work in $\mathbb{Z}$. In particular, if one can show that there are solutions in $Z_p$, then supposedly that implies that there's no elementary proof of the non-existence of solutions in $\mathbb{Z}$. The overreach is the restriction as to what qualifies as "elementary". $\endgroup$ – quasi May 20 '17 at 0:34
  • 1
    $\begingroup$ You haven't dealt with a lot of amateur mathematicians, have you? A lot of people without math sophistication try to solve Fermat. While most people with math knowledge won't make these sorts of errors, the vast majority of "proofs" of Fermat are by unsophisticated amateurs. $\endgroup$ – Thomas Andrews May 20 '17 at 3:07
  • 2
    $\begingroup$ Actually, I have dealt with many such amateurs. and have seen many flawed elementary proofs of FLT. I haven't seen even one such proof, (ignoring the ones by obvious total cranks), which would be disproved by the $p$-adic interpretation in the linked question. $\endgroup$ – quasi May 20 '17 at 3:16
  • 3
    $\begingroup$ However many such proposed proofs can be invalidated by the much more elementary attack I mentioned in my post: How does the proposed proof specialize to the cases $n=2$ or $n=3$? $\endgroup$ – quasi May 20 '17 at 3:23
1
$\begingroup$

The last theorem of Fermat, due to the efforts of lay people, has the stigma of a perpetual motion machine. But we remember that the same stamp had a meteorite theme ...

A serious mathematician will seriously think before publishing a proof for the general case. But to enter the topic and come up with an elementary proof for the case $ n = 3 $ is an excellent warm-up for the brain with non-zero chances of success.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.