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Sigh, this proof bugs me so much. Compute an approximate distance between nth powers, interpret this as the probability of an integer being an nth power, integ

Sigh, this proof bugs me so much. Compute an approximate distance between nth powers, interpret this as the probability of an integer being an nth power, integrate this probability over the sum x^n + y^n, see that the probability of this being an nth power is also very low. I guess this is close enough for government work, but it's so utterly fallacious. For example, the distance between n^2 and (n-1)^2 is 2n - 1. That "means" that the "probability" of N being a perfect square is about 1/(sqrt(2N - 1)). This probability also goes to zero in the limit as N goes to infinity. Not very quickly, but it does. Does that mean that square numbers don't exist? We have many examples of conjectures being disproved by very large counterexamples: https://www.quora.com/What-is-an-example-of-a-conjecture-that-was-proven-wrong-for-very-large-numbers https://www.quora.com/What-is-an-example-of-a-conjecture-tha...

ykler 9y ago
You forgot to sum over N
- jordigh 9y ago
  Yeah, okay, I messed up. Let's argue in a different way: there are approximately sqrt(N) squares between 1 and N, so let's call sqrt(N)/N = 1/sqrt(N) the "probability" that N itself is square. That also goes to zero as N goes to infinity. If we pick higher powers of N, it goes to zero much quicker.
  - nshepperd 9y ago
    That's not what ykler meant. 1/sqrt(N) is the "probability" that N itself is square. We don't care about that, we want the probability that there is any solution to x^2 = N, which is sum{N=0..∞} (1/sqrt(N)). That does not go to zero. ETA: Well, actually the above is the expected number of solutions, so naturally it diverges because there are (infinite) solutions. A more proper way would be to calculate the probability that there are no solutions, which indeed goes to zero. But the probability of there being a solution, and the expected number of solutions, are the same if it is << 1.
    - deleted 9y ago
      [deleted]
todd8 9y ago
Interesting, and startling to me when I first learned of it, is the probabilistic method [1], now not infrequently used in combinatorics to prove the existence of a solution to some question without constructing it. One proves that the probability of existence is strictly greater than zero. [1] https://www.amazon.com/Probabilistic-Method-Discrete-Mathematics-Optimization-ebook/dp/B017JHMKZ4 https://www.amazon.com/Probabilistic-Method-Discrete-Mathema...
- jordigh 9y ago
  Yeah, that's really different, though. It's essentially a way to simplify counting. But the "converse" of the probabilistic method doesn't work: zero probability doesn't mean inexistence in an infinite set.
pmiller2 9y ago
I have to agree that this argument is utterly unconvincing, even as a reason to believe FLT may be true intuitively. It simply dismisses with a wink and a nod the idea that there might be one single counterexample out there for one single exponent (or, a finite number of counterexamples for a finite number of exponents). In other words, it does nothing to argue against the set of counterexamples being of measure 0. Indeed, until Wiles finally proved FLT, the relative lack of progress on Siegel's conjecture on the infinitude of regular primes would tend to lend support to the view that a counterexample might exist somewhere out there.
- alimw 9y ago
  I haven't read it closely, but it looks to me as if the calculation estimates the expected number of counterexamples rather than the "measure" of them (however you've chosen to define that).
  - mathemancer 9y ago
    The set of possible counterexamples has Lebesgue measure zero. https://en.wikipedia.org/wiki/Lebesgue_measure https://en.wikipedia.org/wiki/Lebesgue_measure
- thaumasiotes 9y ago
  > It simply dismisses with a wink and a nod the idea that there might be one single counterexample out there for one single exponent (or, a finite number of counterexamples for a finite number of exponents). In other words, it does nothing to argue against the set of counterexamples being of measure 0. Those "other words" are substantially weaker than the statement before them. If x^n + y^n equaled z^n for every integer x, y, z, and n, the set of counterexamples to Fermat's last theorem would still be of measure 0. Nothing can argue against the set of counterexamples being of measure 0, because the entire problem space itself has measure 0, and the set of counterexamples is necessarily a subset of the problem space.
nshepperd 9y ago
> Compute an approximate distance between nth powers, interpret this as the probability of an integer being an nth power, integrate this probability over the sum x^n + y^n, see that the probability of this being an nth power is also very low. You're mixing up "N is an nth power" with "there exists an N that is an nth power". The purpose over summing over all the integers is to obtain the latter from the former, and it's not necessarily very small even if the former is small.

ykler 9y ago

You forgot to sum over N

jordigh 9y ago
Yeah, okay, I messed up. Let's argue in a different way: there are approximately sqrt(N) squares between 1 and N, so let's call sqrt(N)/N = 1/sqrt(N) the "probability" that N itself is square. That also goes to zero as N goes to infinity. If we pick higher powers of N, it goes to zero much quicker.
- nshepperd 9y ago
  That's not what ykler meant. 1/sqrt(N) is the "probability" that N itself is square. We don't care about that, we want the probability that there is any solution to x^2 = N, which is sum{N=0..∞} (1/sqrt(N)). That does not go to zero. ETA: Well, actually the above is the expected number of solutions, so naturally it diverges because there are (infinite) solutions. A more proper way would be to calculate the probability that there are no solutions, which indeed goes to zero. But the probability of there being a solution, and the expected number of solutions, are the same if it is << 1.
  - deleted 9y ago
    [deleted]

todd8 9y ago

Interesting, and startling to me when I first learned of it, is the probabilistic method [1], now not infrequently used in combinatorics to prove the existence of a solution to some question without constructing it. One proves that the probability of existence is strictly greater than zero. [1] https://www.amazon.com/Probabilistic-Method-Discrete-Mathematics-Optimization-ebook/dp/B017JHMKZ4 https://www.amazon.com/Probabilistic-Method-Discrete-Mathema...

jordigh 9y ago
Yeah, that's really different, though. It's essentially a way to simplify counting. But the "converse" of the probabilistic method doesn't work: zero probability doesn't mean inexistence in an infinite set.

pmiller2 9y ago

I have to agree that this argument is utterly unconvincing, even as a reason to believe FLT may be true intuitively. It simply dismisses with a wink and a nod the idea that there might be one single counterexample out there for one single exponent (or, a finite number of counterexamples for a finite number of exponents). In other words, it does nothing to argue against the set of counterexamples being of measure 0. Indeed, until Wiles finally proved FLT, the relative lack of progress on Siegel's conjecture on the infinitude of regular primes would tend to lend support to the view that a counterexample might exist somewhere out there.

alimw 9y ago
I haven't read it closely, but it looks to me as if the calculation estimates the expected number of counterexamples rather than the "measure" of them (however you've chosen to define that).
- mathemancer 9y ago
  The set of possible counterexamples has Lebesgue measure zero. https://en.wikipedia.org/wiki/Lebesgue_measure https://en.wikipedia.org/wiki/Lebesgue_measure
thaumasiotes 9y ago
> It simply dismisses with a wink and a nod the idea that there might be one single counterexample out there for one single exponent (or, a finite number of counterexamples for a finite number of exponents). In other words, it does nothing to argue against the set of counterexamples being of measure 0. Those "other words" are substantially weaker than the statement before them. If x^n + y^n equaled z^n for every integer x, y, z, and n, the set of counterexamples to Fermat's last theorem would still be of measure 0. Nothing can argue against the set of counterexamples being of measure 0, because the entire problem space itself has measure 0, and the set of counterexamples is necessarily a subset of the problem space.

nshepperd 9y ago

> Compute an approximate distance between nth powers, interpret this as the probability of an integer being an nth power, integrate this probability over the sum x^n + y^n, see that the probability of this being an nth power is also very low. You're mixing up "N is an nth power" with "there exists an N that is an nth power". The purpose over summing over all the integers is to obtain the latter from the former, and it's not necessarily very small even if the former is small.