Feynman on Fermat's Last Theorem (2016)

Is the "past" link being deprecated or something?

I feel like it is a very "physics" motivated approach to at least "investigating" this theorem. Calculating probabilities is a line of thinking Feynman would be familiar with (quantum mechanics). Physics is often responsible for mathematical development, while they are different, they complement each other. It's nice to see different perspectives, and how ideas are connected.

> > Feynman concluded: “for my money Fermat’s theorem is true”.

> > "the main job of theoretical physics is to prove yourself wrong as soon as possible."

> Great example of the main difference between mathematicians and theoretical physicists.

Actually, I'm not sure I agree: even before Wiles's proof, almost every mathematician would have been willing to wager, at least conversationally, on the truth of FLT; and mathematicians also are in the business of proving themselves wrong as soon as possible. The only catch is that we don't count an inability to prove yourself wrong as a proof that you're right ….

This isn't a proof.

By Feynman's argument, you can prove that square numbers almost certainly keep on existing.

Roughly, it goes as such:

1) the probability of N being a perfect square is proportional to 1/sqrt(N).

2) For any N_0 arbitrarily high, if you integrate from N_0 to infinity the expression (1/sqrt(N) dN), you get infinity.

3) The expression in 2) is the "Feynman equivalent" of the expected number of square numbers above N_0.

So Feynman's nonproof actually turns out to be true, despite it not being a proof in this case as well.

No one is claiming that this kind of argument constitutes a proof, but this kind of back of the envelope calculation is very powerful. For example in the theory of prime numbers there is a heuristic that says that primes roughly behave like randomly selected numbers where Prob(N is prime) = 1 / log(N) [this is a simplification, but that's the crux of it]. With this heuristic you can accurately predict whether a large class of statements about primes are true or false, and can get extremely precise estimates about things like "how many twin primes are there less than N", or "how many solutions in primes are there to p1 + p2 = 2*N, for some huge N", which is one way of phrasing two famous prime number conjectures.

It doesn't lead you directly proof - but often just knowing what the answer 'should be' can be a real guiding light.

This argument doesn't calculate the probability of an individual number, it sums the probability over every number.

This kind of thing is used a lot in number theory to figure out the plausibility of some theorem. A lot of open number theory problems are of the form "Prove [unlikely event] never happens."

Edit: Removed braindead argument.

I recall reading that there is an, incorrect, proof that would match the kind of proof we expect from Fermat and thus is believed to be the one he had in mind. However I was unable to find it. Instead I found a discussion on how likely it is that he had a proof : https://hsm.stackexchange.com/questions/3/what-evidence-is-t...

Some even speculate Fermat didn't write this note. Most of his works were collected by his son who is suspected of adding this commentary to a conjecture that was considered many times by Fermat.

I had a number theory professor who said there are a number of deceptively promising avenues of attack using simpler methods than Wiles, but that end up being dead ends. He said mathematicians are now mostly of the opinion that Fermat was probably onto one of these.

Well, I'm not 'math folk' (grey beard programmer), but I love the Horizon documentary on Andrew Wiles and his solution, and I'd love to hear why my intuitive understanding is inapplicable from people who know better than myself. (Note that this will not in any way be a proof, but just the train of thought I believe to be the line of thinking Fermat may have used to construct his proper mathematical proof.)

My idea here is based upon physical/visual intuition, starting with why it works for n=2 (squares) and then why it cannot work for n=3 (cubes) and then that n>3 is necessarily more complex than n=3 thus cannot work either.

[Note that I will use lower case letters for the sides/roots and their uppercase letters to denote the areas or volumes. Thus, the full equation is Z=Y+X, with X = x^n, resulting in z^n = y^n + x^n. I also use (for n=2), dy = z - y, and Dy = 2(y(dy)) + dy^2, and dx = z - x, and Dx = 2(x(dx)) + dx^2. I'm sorry my dx and dy conflict with calculus notation but my notation means dx is "the difference between z and x" which is the same as "the length that must be added to x to equal z" and Dx is "total amount that must be added to X to get Z". Therefore (for n=2), Dx = Y = 2(x(dx)) + dx^2, and Dy = X = 2(y(dy)) + dy^2. ]

For n=2, Z=Y+X works because (what can be visualized as a square) X can be "smushed" over two sides and their joining corner of (the other square) Y evenly, such that Z = Y + Dy = Y + 2(y(dy)) + dy^2. The term "2(y(dy))" is the amount that must be added along each of the two sides, and the term "dy^2" is the amount that must be added at the corner to complete the perfect square Z.

So, for example, 5^2 = 4^2 + 3^2 because both 3^2 = 9 = 2(4(1)) + 1^2 = 2(4) + 1 = 8 + 1, and 4^2 = 16 = 2(3(2)) + 2^2 = 2(6) + 4 = 12 + 4.

Now, for n=3, we must visualize the situation where the cube X is smushed over the cube Y's three faces and its joining corner. (Now X=x^3 and Y=y^3.)

The equations for dx and dy are the same, but Dx and Dy have expanded by a dimension: Dx = Y = 3(x^2)(dx) + 3(x)(dx^2) + dx^3, and likewise Dy = X = 3(y^2)(dy) + 3(y)(dy^2) + dy^3.

The term "3(x^2)(dx)" is the amount that must be added to three faces of the cube Y, the term "3(x)(dx^2)" is the amount that must be added along the three edges joining those three faces of the cube Y, and the term "dx^3" is the amount that must be added at the corner.

Now, I haven't the maths to prove why Dx and Dy for n=3 won't have integer solutions but my intuition says it has something to do with the fact that it's three dimensions and, therefore, a couple of odd numbers multiplying around in there (the first two terms) and the fact that there are only two cubes being smushed together to try and reconstitute another perfect cube. I also imagine Fermat could actually mathematically prove why it's impossible. Perhaps it can be shown that Dx and Dy cannot both have diophantine solutions. These are just guesses.

As for n>3, the terms (and physical/visual representations) will only become more complex and there will be still only two terms with which to reconstitute the hypercube.

Anyway, that's my intuition about the entire problem and I have to imagine that a proof that Fermat can easily intuit yet is (a bit?) too large to fit in the margin must surely tread down a simple path, perhaps even one that relies on a physical/visual interpretation of what the equations can be likened to.

I look forward to this being eviscerated or flatly rejected, if appropriate, or at least corrected for inconsistencies. If it serves anyone in their exploration of this insidiously complex yet apparently simple-seeming problem, my joy would only grow. If my name would someday appear in a mathematical paper that a real mathematician produces as a result of this, well that would be out of this world for this poverty-striken math wannabe.

[Edited to fix my n=3 equations.]

> It's not that hard people. Stop believing everything you're told about how "hard" something is.

There are still many problems in physics and mathematics which are considered "hard" (e.g., dark energy, Riemann hypothesis, etc). Can we crack them by simply adopting your positive mindset?

Sounds interesting, but not sure what you mean by:

> you'll discover that there will always be a gap if you try and combine two odd number series together

Can you elaborate?

Off topic: why doesn't HN support LaTeX?

Do you really believe that:

(a) This constitutes a proof;

(b) This is the "proof" that Fermat had;

(c) Mathematicians missed this for over 350 year?

I'm not quite sure exactly what you are claiming.

> The same trick works for higher powers.

can you demonstrate?

I'm working to understand this, but I can't seem to fit it together. Following Feynman's lead in this sort of thing, can you give me an explicit example of why the equation x^13+y^13=z^13 has no solutions? Or even just use your technique to explain why x^5+y^5=z^5 has no solutions?

Thanks.

Can you please read the site guidelines and follow them?

https://news.ycombinator.com/newsguidelines.html

This argument puzzled me, but unfortunately this is as far as I got:

~~Lemma 1~~:

    z^n can be written as a difference of squares:

Proof:

z^n = x^2 - y^2 = (x+y)(x-y), leading to the system of equations

x + y = z^(n-1)

x - y = z

which may be solved for x and y:

x = (z^(n-1) + z)/2

y = x - z

QED.

~~Lemma 2~~:

    Any number squared can be written as a sum of sequantial odd numbers starting at 1.

By induction:

(n)^2 = (n-1)^2 + (2(n-1)+1) = \sum_{k=0}^{n-1}(2k + 1)

QED.

~~~Fermat's Last Theorem~~~:

    z^n = x^n + y^n has no solutions for n>2, and positive z, x, y.

Proof:

Without loss of generality, assume a > b, then rewrite z^n as a sum of sequential odd numbers starting at b:

z^n = a^2 - b^2 = \sum_{k=b}^{a-1}(2k+1)

x^n and y^n can similarly be written as a sum of sequenatal odd numbers:

x^n = c^2 - d^2 = \sum_{k=d}^{c-1}(2k+1)

y^n = e^2 - f^2 = \sum{k=f}^{e-1}(2k+1)

By substitution to the Theorem's equation:

\sum_{k=b}^{a-1}(2k+1) = \sum_{k=d}^{c-1}(2k+1) + \sum_{f}^{e-1}(2k+1)

Which is true if and only if there are no gaps in the bounds of summation on the right side, so d = b, c = f, and e = a. But then

a^2 - b^2 = (f^2 + (-a^2))^n + (b^2 + (-f^2))^n

and this is certainly not true by the binomial theorem. We've reached a contradiction so QED.

^^ This is where I'm stuck. I don't actually know why that would not be true by the binomial theorem? It seems like simple expansion should do it.