
Feynman on Fermat's Last Theorem - bezierc
http://www.lbatalha.com/blog/feynman-on-fermats-last-theorem
======
sajid
If FLT hadn't already been proven then Feynman's argument would explain away
the substantial numerical evidence collected in it's favor. In other words, it
would suggest that FLT is just a statistical accident and true for no
particular reason.

Of course, we now know that FLT is related to deep ideas in number theory.

~~~
qmalzp
I mean Feynman's argument really seems to show that it should hold for
sufficiently large N. I think one could reasonably argue that it's a
statistical accident that FLT holds for N >= 3 as opposed to say, N >= 100. I
think even Wiles' proof only works for N sufficiently large (at least N >= 5).
Small N of course were handled by earlier results.

------
ahh
Very cute argument, and very much in his style--he was famous, as the author
notes, for heuristic arguments that weren't very formalizable but had a lot of
beauty.

One story I heard is that a computer scientist tried to explain the P=NP?
problem to him; Feymnan couldn't understand why this was a problem. It was
obviously true that P != NP, what even needed proving?

~~~
bashinator
Per "Surely You're Joking" and the article, one of the cognitive techniques
Feynman used was to keep a physical example, a demo, in his mind that
conformed to the math being explained. I wonder if he used that for this, and
what his model was.

~~~
jsprogrammer
If you imagine it as a logical statement, represented physically in writing: P
<=> ~P

Then, I think it is clear the statement is not true. NP is trivially
definitionally not equivalent (equal) to P.

~~~
Chinjut
NP is most certainly not "trivially definitionally not equivalent (equal) to
P". (Do you think everyone puzzling over the question of whether P = NP is a
moron? You are aware there is a million dollar prize for a proof one way or
the other, yes?). What definition are you using for NP?

~~~
jsprogrammer
P=deterministic algorithms in polynomial time

NP=non-deterministic algorithms in polynomial time

See, it is a definitionally trivial logical negation.

 _That 's the joke!_

Edit: Also, why cannot I report/flag your post? Your personal malignment is
uncalled for.

~~~
Chinjut
It's true I did not understand you were joking, but I didn't personally malign
you either. If you are referring to the word "moron", I didn't call you a
moron; I was shocked that you thought P vs. NP was so trivial, and asked
whether you therefore thought everyone puzzling over it was a moron.

~~~
jsprogrammer
I didn't say you maligned me. You implied others were morons (or, at least
that I would think that).

You should be directing your disbelief at Feynman's remains. He is the one
that made the claim. I just explained the joke.

Anyway, it is still true that the physical model of P != NP is definitionally
trivial.

~~~
Chinjut
I didn't malign others, either, as I didn't call anyone a moron... Yes, it's
true I implied that you might think that others who puzzled over P vs. NP were
morons. I was, after all, laboring under the belief that you thought P vs. NP
was trivial. Is that a malignment?

Anyway. Nevermind that. Feynman joked that P was trivially not equal to NP?
Where can I find that joke given by Feynman?

~~~
jsprogrammer
>Where can I find that joke given by Feynman?

Approximately 6 parents up this comment tree.

I'll address your other comment here (HN has an inexplicably bad rate limit):

The physical model is the literal characters:

P != NP

:)

~~~
Chinjut
I mean a cite that Feynman made that joke; where can I find that?

~~~
bashinator
This whole talk is about it

[https://www.youtube.com/watch?v=v3pYRn5j7oI](https://www.youtube.com/watch?v=v3pYRn5j7oI)

------
mark-r
The last quote on the page should be familiar to HN readers as a different
version of "fail fast, fail often".

~~~
theseatoms
In other words, prefer breadth-first searches to depth-first searches.

~~~
semi-extrinsic
Mandatory [https://xkcd.com/761/](https://xkcd.com/761/)

------
conistonwater
1e+33 is not such a big number as far as number theory goes (for example:
[http://mathoverflow.net/questions/15444/examples-of-
eventual...](http://mathoverflow.net/questions/15444/examples-of-eventual-
counterexamples),
[https://en.wikipedia.org/wiki/Skewes%27_number](https://en.wikipedia.org/wiki/Skewes%27_number)).
It's still a nice exercise.

As another example, consider the question of whether there exists a right-
angled triangle with rational sides, having an area of 157.

~~~
acqq
If I understand it right, it's a number with which he could have estimated if
it's worth using a computer to perform a brute force search for a possible
solution.

Once he had 1e-33 for _all_ n > 100 that could mean that even trying with 1M
computers where each tries 10G solutions per second (1e6*1e10) some millions
of years could pass without the positive result. Then it's exactly reasonable
to say "for my money Fermat’s theorem is true" as in, really not worth trying
blindly.

~~~
conistonwater
I can't believe that the right interpretation of what he said. He must have
known what the actual FLT meant. Furthermore, it would go against the
mathematical tradition of what it means to "think that a conjecture is true".
Compare with RH, for example: when people say it's probably true they
absolutely do _not_ mean it's true for all small numbers, in part because
there are built-up areas of mathematics that depend on it being exactly true.
Besides, brute-force search is a pretty terrible algorithm (in general), so
finding out that it fails on a particular problem isn't that interesting.

~~~
acqq
He died in 1988.

Imagine somebody came to you at these "early" times (Wiles proved the theorem
in 1995) with a "grand project" to use a lot of computers to search for a
possible "solution," being able to try 10 billion Ns in one second. What would
be your argument _against_ such a project? Would you try some similar
derivation to get an estimate of success?

I know, brute forcing _all integers_ is impossible, but you can even "imagine"
a "superquantum (and now not existing) computer" which can do a lot of small
integers in parallel.

And yes, I know, probabilities aren't proofs, especially not for integers. One
counterexample is enough:

3987^{12} + 4365^{12} = 4472^{12}

Also interesting to see the accidental(?) order of magnitude of the numbers
involved.

------
mrcactu5
I have never seen a probabilistic approach to FLT but there is a branch of
math called Probabilistic Number Theory

[https://people.math.ethz.ch/~kowalski/probabilistic-
number-t...](https://people.math.ethz.ch/~kowalski/probabilistic-number-
theory.pdf)

I am also not 100% sure Feynman wrote something like this. However the 2014
Fields Medal was awarded to Manjul Bhargava for studying random elliptic
curves.

[https://www.quantamagazine.org/20140812-the-musical-
magical-...](https://www.quantamagazine.org/20140812-the-musical-magical-
number-theorist/)

------
erikbern
There's a similar argument for why Goldbach's conjecture* might just be a
statistical fluke. I remember realizing this in high school and deriving the
probability – alas it's not a novel idea and Wikipedia has a great summary:
[https://en.wikipedia.org/wiki/Goldbach%27s_conjecture#Heuris...](https://en.wikipedia.org/wiki/Goldbach%27s_conjecture#Heuristic_justification)

It's very possible that it's an unprovable true conjecture, which is kind of
interesting in itself.

* every even number can be written as a sum of two primes

~~~
Houshalter
What's interesting about this argument is, if its true, it's an example of a
simple conjecture that is true but unprovable. But not like Godel sentences,
which can be proved under different axioms, or proved to be unprovable. This
would be completely unprovable. And not because of some logical paradox, but
simply because there is no mathematical reason for it to be false, it's just a
coincidence.

There are also probably many conjectures like this. Freeman Dyson constructed
one, that no digits of powers of two, reversed, make a power of 5. The
reasoning being that the digits grow big quickly, and the base 10
representation is basically random and uncorelared with it.

I think other conjectures about digits of numbers are similar. E.g.
Mathematicians have found it impossible to prove that any numbers are normal,
or even if pi contains infinite 5's, and many similar properties. These things
may only be probabilistically true.

~~~
tgb
Well, any statement can be proven under different axioms. Just take that
statement itself as your only axiom.

------
joshdick
I wonder how many false conjectures could pass muster using this sort of
probabilistic argument.

~~~
Someone
I guess every false conjecture can be made to pass it. The trick is to make
the set of items searched in large enough.

For example, to show that no elephants exist, start with the (infinite) set of
all possible chromosome sets. The proportion of them that produces an elephant
is zero. QED.

Examples from mathematics:

The number 42 does not exist (logic: pick an integer. The probability that it
equals 42 is zero. QED)

There are no even primes.

There are no primes.

There are no integers.

There are no rational numbers.

All numbers are transcendental.

Continuous functions do not exist.

There are no regular polygons.

~~~
Analemma_
Whoa, hang on. Statistical arguments for unproven conjectures are bad, but
this counterargument is as bad or worse, especially when you start talking
about infinity. Just to address your first example:

> The number 42 does not exist (logic: pick an integer. The probability that
> it equals 42 is zero. QED)

I object! What is your probability distribution function over the integers?
Your phrasing sort of implies a uniform distribution, but there is no such
thing as a uniform distribution on an infinite set, and as soon as you pick a
plausible pdf the argument stops working.

~~~
jordigh
It is easy to address your objection. On a uniform distribution on [n] := {0,
1, 2, ...n}, P(X=42)->0 as n->infty. This is similar to Feynmann's argument.

~~~
tgb
You're forgetting to integrate the probability over the domain 0 to n. If you
do, you always get 1 for n at least 42.

~~~
jordigh
Those are two separate events. I'm talking about the event n = 42, which makes
sense for all [n]. You're talking about the event n >= 42, which is a very
different event.

~~~
tgb
No, I'm not talking about integrating the probability that n >= 42.

What's the chance that a random number in [n] is 42? It's 1/n, if n is at
least 42. Sum that over all x in [n] and you get 1, i.e. the probability that
there is a number in [n] that is 42.

~~~
jordigh
You're still talking about different events. The event that there is 42 in [n]
is different from the even that x = 42.

~~~
tgb
Then I guess I just don't understand what you're trying to conclude from this
argument.

~~~
jordigh
Richard Feynmann was trying to argue probabilistically about the possible
existence of a counterexample to FLT. To do this, he more or less tries to put
a probability value on the event "a counterexample exists". The problem is
that his kind of reasoning is similar to the probability of an event such as
"x = 42". Counterexamples to FLT may, a priori, be as rare as the number 42 is
within the integers, i.e. with probability tending to zero as you consider
larger and larger sets of integers. But a very small probability of a
counterexample, even a probability that tends to zero, does not mean a
counterexample doesn't exist anymore than it says that 42 doesn't exist.

~~~
tgb
See I can't see the analogy from your argument to Feynman's for the following
reason: you are computing the probability that x=42 for an individual x and
saying that it is going to 0 as x->infinity. Feynman first computes the
probability that N is x^n + y^n but then _integrates that over all N and n_ in
order to get the probability that _there is a solution anywhere_. His argument
takes into account the fact that there could be just one single lone solution
in a infinite sea of non-solutions, since his probability estimate is for
"there is a solution _anywhere_ " not "this particular number is a solution".
And I think that when you add that part to your estimate, you do get a
probability of 1 that 42 exists somewhere, even if individually you get a low
probability of it being at any given large x.

~~~
jordigh
This detail doesn't matter. He's still computing a tiny nonzero probability
that is only zero in the limit. We came up with other examples of tiny nonzero
probabilities that are zero in the limit that can lead to the wrong
conclusion, such as saying that 42 doesn't exist.

Feynmann's argument is clumsy and wrong. This kind of approximation is ok for
a physicist but it doesn't work for mathematics.

------
utopcell
By this probabilistic reasoning, we shouldn't bother searching for
"meaningful" polynomial time algorithms because they are practically non-
existent in the sea of all exponential time algorithms. If anything, this
argument should be a motivator to work on a problem because if an underlying
structure exists, it would be rare and beautiful. Very nicely written article,
showing us a glimpse of the inner workings of a great mind.

------
ambrop7
"the probability that N is a perfect n^nth power..."

Can someone explain what probability means here in relation to N? From my
understanding, it depends on what N is for you. If it's a constant, that
probability is obviously 0 or 1. So that can't be it. Then N must be some kind
of random variable. But with what distribution? And in what kind of system can
the probability of event(random_variable) involve random_variable itself?

~~~
lsh123
The meaning here is: pick a positive integer N, what is the chance (aka
probability) of it being an n-th power of another positive integer. And you
are correct, this probability depends on N.

~~~
ambrop7
I don't get it. The probability must depend on how likely I am to select any
specific integer, i.e. probability mass function of N. The probability cannot
depend on the value of the random variable itself but could involve any
parameters that define its distribution.

I would consider something like the following a valid question: "Let N be a
random integer between 0 and M-1 with uniform distribution. What is the
probability that N is even?"

Then an answer could be "the probability that N is even is 1/2 if M is even
and (M+1)/(2M) if M is odd". See this does not involve N but does involve M
which is a parameter for the distribution of N.

Your explanation seems to invoke some "common sense" which I am not able to
unify with my understanding of probability theory.

~~~
ikeboy
One meaning is that summing the probabilities for N=1,2,3, ... and checking
how many numbers actually meet the criteria yield the same result in the
limit. More precisely, their ratio goes to 1 as N goes to infinity.

Edit: define f(N) as the number of numbers below N that match. Consider the
function f(N)/N. We call a function of N the probability of N matching if it
asymptomatically approaches f(N)/N.

------
ethan_g
Apologies to rain on the parade, but this sort of argument is very well known
and commonly used in number theory. Since these techniques don't lead to
proofs, they may not get as much attention in standard math education, but
they are still well-known. Many number theory papers use this technique (often
with more refinements) to estimate the number of expected solutions. I would
be stunned if it wasn't used on Fermat's last theorem long before Feynman did.

For an exposition on some heuristics mathematicians use,
[https://terrytao.wordpress.com/2015/01/04/254a-supplement-4-...](https://terrytao.wordpress.com/2015/01/04/254a-supplement-4-probabilistic-
models-and-heuristics-for-the-primes-optional/) is a good read (although it's
aimed at readers already having some background in number theory).

------
Someone
Using way simpler math, the probability that an even number is prime is zero.
One cannot conclude from that that there are no even primes (in fact, many
mathematicians think the smallest prime is even
([https://cs.uwaterloo.ca/journals/JIS/VOL15/Caldwell2/cald6.h...](https://cs.uwaterloo.ca/journals/JIS/VOL15/Caldwell2/cald6.html)
))

Mathematicians do use this kind of back of the envelope calculations to get a
feeling for whether a statement may be true, but they can never prove
something.

~~~
jakeva
Your link doesn't work for me…

~~~
nappy-doo
Remove trailing ):

[https://cs.uwaterloo.ca/journals/JIS/VOL15/Caldwell2/cald6.h...](https://cs.uwaterloo.ca/journals/JIS/VOL15/Caldwell2/cald6.html)

------
lmeunier
I don't understand why the distance between root(n, N) and root(n, N+1) is the
probability that N is a perfect nth power. What if we choose to compute the
distance between root(n, N) and root(n, N+0.5) ? By the same logic, if N is a
perfect nth power, there exists one integer in the interval [root(n,
N),root(n, N+0.5)], and since the distance between all consecutive integers is
one, the probability is the ratio etc ... And what about 0.2? 0.1? We can keep
going until the probability becomes zero ! And this can't be !

------
raverbashing
It's a nice exercise

However Number Theory is a different beast altogether

It has as much to do with "regular" math as English and Latin have in common,
even though they are written with the same alphabet.

~~~
siegelzero
That's not exactly true. Density arguments similar to this are fairly common
in number theory (even if the arguments aren't rigorous). Granted, one has to
be careful with arguments like this, but they can often be useful.

------
plg
i have to say this never-ending fixation / worship of Feynman is a bit creepy

~~~
lb1lf
There's hardly any denying that Feynman was quite a character; he (sometimes
spectacularly) failed to live up to our expectations as to how such a
brilliant intellect ought to behave.

That is probably a significant reason for his enduring appeal; a streetwise,
bongo-playing physicist with a talent for quips.

~~~
xiaoma
>"he (sometimes spectacularly) failed to live up to our expectations as to how
such a brilliant intellect ought to behave."

Can you elaborate on this?

------
supergirl
finding that the probability is extremely small doesn't really get you any
closer to proving it. for such a hard to prove theorem, it makes sense the
probability is small. that's why it was interesting in the first place.

I find it hard to believe Feynman concluded from this that the theorem is
probably correct. it only takes 1 case among an infinity to make the theorem
false.

~~~
Retra
I think the idea here is that, in general, simple coherent properties of
number tend to show themselves on smaller numbers more readily than on larger
ones.

It would be quite unusual to have a problem stated in such simple terms and
require a solution so far away. Thus this probability is used to infer just
how large such a number must be, and eventually you're going to hit a point
where there is more information encoded in the number than there is necessary
to solve the problem, as which point no larger number could be a solution.

This is basically how induction works anyway: you produce a base case and an
algorithm, and infer that the information contained therein meets the needs of
the problem. Then any number which would encode more information is irrelevant
(and thus sufficient) and you have an inductive proof.

~~~
kryptiskt
Littlewood's prime counting theorem is an example what can hide up among the
big numbers
([https://en.wikipedia.org/wiki/Skewes%27_number](https://en.wikipedia.org/wiki/Skewes%27_number))

------
fareesh
My math is passably college level, so most of these proofs go over my head.
I've always thought of FLT as stating "You cannot make n-3 cubes of side z by
stacking/tiling n-3 cubes of side x and y", at which point it becomes
something of a tessellation problem in 3 dimensions, which would seem to me to
be the intuitive way to go about proving this.

~~~
Chinjut
Sorry, I'm having difficulty following you. What do you mean by an "n-3 cube"?
Or are you actually talking about n - 3 many different cubes?

~~~
fareesh
My bad - I meant a^(n-3) many different cubes of side a plus b^(n-3) many
different cubes of side b needing to cover the total volume occupied by
c^(n-3) many different cubes of side c. In other words it is a way to try and
find a proof by thinking in terms of shapes and using some kind of tesselation
method.You could do the same in squares too I suppose.

~~~
Chinjut
Or just in lines... a^n many lines of length 1 + b^n many lines of length 1 to
cover the total length of c^n many lines of length 1.

Or just in n-dimensional cubes from the start: 1 n-dimensional cube of side-
length a + 1 n-dimensional cube of side-length b to cover the volume of 1
n-dimensional cube of side-length c.

Of course, both of these don't amount to much different than just saying a^n +
b^n = c^n directly. :)

I don't suspect that decomposing it into a^(n - k) a^k + b^(n - k) b^k = c^(n
- k) c^k for arbitrary k is particularly helpful, and, for reasons as
illustrated above, I suspect the aid to intuition from thinking in terms of
volume and tessellations geometrically isn't very great, but, I'm not an
expert on Fermat's Last Theorem. (For all I know, something like this does get
used in the proof...)

------
kenjackson
I don't understand what "the probability of" means here. Does the probability
of 10^-31 mean that if we construct the integers in a different way, it might
result in the theorem being false? Or is it that if you change the equation
there is only at 10^-31 probability that you arrive at a theorem which is
false/true? Or something else altogether.

~~~
acqq
See my other comment here for how I understood it. Note "for any N > N0 = 100"
in the article.

[https://news.ycombinator.com/item?id=12019762](https://news.ycombinator.com/item?id=12019762)

------
cantrevealname
> _Feynman also knew about Sophie Germain’s result, who proved in the early
> 19th century that Fermat’s equation has no solution for n≤100._

That's intriguing! What's special about the number 100 that you can prove the
case for n≤100?

~~~
Chinjut
That description of what Sophie Germain did is slightly glib and inaccurate, I
believe. I will give an abbreviated, but accurate, to the best of my
knowledge, explanation of what happened:

Note that, if you have proven Fermat's Little Theorem for some exponent, then
you have also proven it for every multiple of that exponent. Thus, to prove it
for all exponents (> 2) up to some limit, it suffices to prove it for all odd
prime exponents up to that limit, as well as for exponent 4. Fermat himself
gave an argument which worked for exponent 4, so afterwards, one could
consider only odd prime exponents.

Note also that, if p is prime, and we have a solution to x^p + y^p = z^p, then
out of {x, y, z}, precisely 0, 1, or all 3 are divisible by p (i.e., if any
two were divisible by p, then so would be the third). If indeed all 3 are
divisible by p, we may accordingly divide all through by p to obtain a smaller
solution; thus, if there is any solution, there is a minimal solution where
precisely 0 or 1 of {x, y, z} are divisible by p.

So to prove Fermat's Little Theorem for prime exponent p, it suffices to prove
both of the following claims for all solutions to x^p + y^p = z^p:

(A) It cannot be the case that precisely 0 of {x, y, z} are divisible by p

(B) It cannot be the case that precisely 1 of {x, y, z} is divisible by p

If this can be done for every odd prime p up to some limit, FLT is established
for all exponents up to that limit.

Germain did not do this. She did not have a strategy for proving (B). This
prevented her from establishing FLT in full for any exponent.

Germain did manage, however, to discover a strategy for establishing (A) for
various p. Specifically, she discovered a sufficient condition for (A) was the
existence of another prime t in a certain decidable relation to p. Germain
then manually searched for and discovered such t for each prime p < 100.

Why'd she stop there? Because it seemed like a nice place to stop. Nothing
special about 100 except the human factor. One could keep going, and indeed,
Legendre extended Germain's results to each prime < 197\. Germain and Legendre
were both aided by certain techniques they had developed in order to quickly
identify certain candidate t that could work as the auxiliary primes for
particular exponents p; however, these techniques broke down at 197, and while
it is possible to find by brute force an auxiliary prime t in the appropriate
relation to p = 197, the smallest such t is very large and the scale of the
calculation was beyond feasible at the time.

I may still not have gotten the story exactly correct, or noted all the
pertinent details, but for more, see
[https://www.agnesscott.edu/lriddle/women/germain-
FLT/SGandFL...](https://www.agnesscott.edu/lriddle/women/germain-
FLT/SGandFLT.htm), on which I based the above description.

~~~
Chinjut
A little more detail, since I bothered to go learn it:

Lemma: If x and y are coprime, then the gcd of x + y and (x^n + y^n)/(x + y)
divides n.

Proof: Expand out the polynomial division (note that x + y does indeed divide
x^n + y^n), and then divide the result by (x + y) again, observing a remainder
of ny^(n - 1). Thus, the gcd of interest is the same as gcd(x + y, ny^(n -
1)). As y^(n - 1) is coprime to x + y (by coprimeness of x and y), this is
furthermore the same as gcd(x + y, n), completing the proof.

Lemma: If x^p + y^p + z^p = 0, for odd prime p, with z indivisible by p, then
(x + y) and (x^p + y^p)/(x + y) are coprime p-th powers.

Proof: As z is indivisible by p, so is -z^p = x^p + y^p, and thus so is its
factor x + y. At this point, invoking the above lemma and the primeness of p,
we find that (x + y) and (x^p + y^p)/(x + y) are coprime; as their product is
a p-th power (-z^p), we conclude they are furthermore each p-th powers.

Sophie Germain's theorem: Suppose p is an odd prime, t is a prime, and the
mod-t exponent-p case of FLT is true (in the sense that there is no solution
to x^p + y^p + z^p = 0 (mod t) where x, y, and z are nonzero (mod t)), but the
non-modular exponent-p (A) case of FLT fails (in the sense that there IS some
solution to x^p + y^p + z^p = 0 in integers, all indivisible by p, which we
can assume minimal so that x, y, and z are pairwise coprime). Then p is a p-th
power modulo t.

Proof: Invoking the above lemma, we have some a, b, c such that a^p = y + z,
b^p = x + z, and c^p = x + y.

Furthermore, since mod-t FLT is true, we have that (precisely) one of x, y, or
z is zero mod t; WLOG, let this be x. But as 2x = b^p + c^p - a^p, we can
invoke mod-t FLT again to conclude that one of b, c, or a is zero mod t. It
cannot be b or c, as y and z, respectively, are nonzero mod t; thus, a is zero
mod t. Thus, y + z = 0 (mod t), and therefore, by expanding out the polynomial
(y^p + z^p)/(y + z), we see that its integer value is equal to py^(p - 1)
modulo t. As x is zero mod t, we must have furthermore that (x^p + y^p)/(x +
y) must be y^(p - 1) modulo t (note that this is nonzero modulo t). As the
former and latter are both p-th powers by the above lemma, so is their ratio
in modulo t arithmetic, which is p, completing the proof.

We can now rephrase Sophie Germain's theorem like so: To establish the (A)
case of FLT for exponent p, it suffices to find some prime t such that BOTH
the mod-t exponent-p case of FLT is true AND p is not a p-th power modulo t.

Such a t is the auxiliary prime for p discussed above. Note that the truth or
falsehood of this condition on t relative to p is decidable by finite search.

A couple more comments: As a consequence of the multiplicative group modulo a
prime being cyclic, we have that, for primes p and t, EVERY value is a p-th
power modulo t unless t is 1 mod p. Thus, an auxiliary prime t for odd prime p
must be of the form kp + 1. Furthermore, as t cannot be 2, it must be an odd
prime, and therefore k must be even. Finally, we can also rule out k divisible
by 3 by again invoking the cyclicity of the multiplicative group modulo t
(were k divisible by 3, we would have some primitive cube root of 1 modulo t
which was furthermore a p-th power; the three powers of this value would sum
to zero and thus provide a counterexample to the mod-t exponent-p case of
FLT). So when searching for auxiliary t to prime p, we are looking for primes
of the form kp + 1 where k is an even value not divisible by 3; for all such k
up through 16, Germain and Legendre managed to prove that conversely, whenever
kp + 1 is prime, it satisfies all the conditions to be an auxiliary prime for
p (with a slight exception for the two cases where p = 3 and k is 14 or 16),
which provides auxiliary primes for each p < 197; however, at p = 197, one
must go beyond this, to a minimal k of 38.

I do not at the moment know if there are auxiliary primes for each p (and I
suspect no one does?), nor how one goes about showing that, for various k, kp
+ 1 is automatically an auxiliary prime whenever prime (or whether this
generalizes beyond k = 16). Oh well. More to learn, always.

------
paulpauper
The immediate problem with this result, upon first inspection, is that it's
not unique to FLT. One can derive other integrals and inequalities that also
yield tiny probabilities.

------
Animats
That approach is so Feynman.

It's too bad he didn't live to see machine learning take off. Feynman thought
statistically, and probably could have advanced the field.

------
Chinjut
This comment isn't meant to be dismissive. I think it's interesting to explore
arguments like this. Math progresses by exploring thoughts. But...

Couldn't we argue in the same way that there are unlikely to be any solutions
to the simpler equation x^n = z^n?

After all, taking p(x) = x^(1/n - 1)/n to be the probability that x is an n-th
power, as Feynman does, and then integrating p(x^n) dx from x = x_0 to
infinity to find the expected number of n-th powers of the form x^n for x >
x_0, as Feynman does with p(x^n + y^n), we find for n > 2 that this comes out
to 1/(x_0^(n - 2) * n * (n - 2)), which is quite small for sizable x_0 and n.

This yields, for example, that the expected number of solutions (and thus an
upper-bound for the probability of the existence of any solutions) to the
equation x^100 = z^100 with x > 10 should be 1 out of 98 googol. This is about
as certain as certain gets that there are no solutions... and yet solutions
are as ubiquitous as ubiquitous gets!

~~~
tvural
Well, Feynman is assuming that x^n + y^n is as likely as any other number to
be an nth power. He shows that under this assumption, FLT is very likely true.
So if FLT is false, it probably wouldn't be due to some coincidental
counterexample. There would have to be a mathematical structure forcing x^n +
y^n to be an nth power in some cases. In your example, the mathematical
structure forcing x^n to be an nth power is obvious.

The downside of Feynman's approach is that it can't get you any intuition
about the structure, the things that aren't statistical randomness. And
whether FLT was true or false depended exactly on whether the structure
pointed one way or the other. So I have no idea why he was so confident in
this analysis.

~~~
Chinjut
Fair enough on your first paragraph. (Though what distinguishes mathematical
facts which are "coincidence" from mathematical facts forced true by
mathematical structure?)

Actually, I'd say it is odd to find this sort of analysis to give great
confidence about the results, not just because it ignores the possibility of
structure, but also because it ignores the possibility of coincidence!

After all, there are some things which we probably want to call mathematical
coincidences which heuristic argument would tell us are bogglingly unlikely
and yet which nonetheless happen. For example, another probabilistic argument:
Consider the basic arithmetic 2-ary operations +, -, * , and ^. There are
20!/(10! * 11!) full binary trees with 11 leaf nodes, 10^4 ways to label their
internal nodes with one of these 4 operations, and 11! ways to assign the
values 0 through 10 to those leaf nodes in some permutation. Thus, there are
at most 20!/10! * 10^4 (overall, less than 10^16) values which can be
generated using these operators and the natural numbers up through 10 once
each (and this is an overestimate, ignoring the structure of, e.g.,
commutativity of + and * which causes less distinct values to be produced).

We would expect, therefore, that the closest we could get one of those values
to line up with a particular unrelated mathematical constant, even focusing
only on the fractional part, is no more than about 16 or so decimal digits. If
there are thousands of mathematical constants we might compare it to, perhaps
we'd get a match to about 20 digits on one.

And yet! And yet (1 + 9^(0 - 4^(6 * 7)))^(3^(2^(10 * 8 + 5))) lines up with e
to 18457734525360901453873570 digits.

Now, what's happening here is that we have this other nice lining up of
9^(4^(6 * 7)) = 3^(2^(10 * 8 + 5)), which we can then plug into e ~= (1 +
1/n)^n with huge n. And this, in turn, is because 9 = 3^2 and 4 = 2^2 and 1 +
2 * 6 * 7 = 10 * 8 + 5. There's a bit of an explanation. And yet... surely if
anything is a coincidence, this is?

So... I guess what I'm saying is, it is odd to use probabilistic arguments to
rule out coincidence. The whole thing that makes a coincidence remarkable is
that it is the sort of thing we would consider unlikely, and yet such things
_do_ occur, even in mathematics.

~~~
tvural
I don't consider the example you gave to be a coincidence, because if you know
the definition of e then you know why one of those numbers could line up
really, really well with e.

I'm basically saying that Feynman's argument rules out seemingly random
counterexamples like 27^5 + 84^5 + 110^5 + 133^5 = 144^5, which disproved the
sum of powers conjecture.

~~~
Chinjut
It seems to Feynman's argument works via the opposite of ruling out
coincidences of that sort. It shows "If nothing surprising happens, then we
expect very nearly zero counterexamples to Fermat's Last Theorem"... But the
"surprising" in "If nothing surprising happens" encompasses both surprising
structure (things which are true for some deep, clean reason) AND surprising
numeric coincidences (things which are true for no good reason, but just
happen, surprisingly, to line up in a nice, unexpected way).

------
klue07
> is far from the real 110 pages long proof of FLT that took A.Whiles years to
> put together

He could've gotten away with the pun "took A. Wile to put together."

~~~
gohrt
It long time, and a lot of Wiles

------
GFK_of_xmaspast
Probabilistic number theory is an established thing, c.f. lots of Erdos's
stuff.

