3 ms·

Pierre de Fermat's Link to a High School Student's Prime Math Proof

24 points by nsoonhui 3y ago

inglor_cz 3y ago
Wow, number theory was once my thing... Carmichael numbers are interesting indeed. Their very existence indicates that the "true nature" of primes is more obscure and complicated than people of the earlier ages thought. Some applies to regular and irregular primes. If all primes were regular, Fermat's Last Theorem would have been a lot easier to prove. It is possible that Fermat himself was mistaken and saw the outlines of the proof for regular primes. Also, maths is one of the last remaining sciences where a fresh insight from a relative newbie can still move major, century-old obstacles.
tromp 3y ago
The author appears to go astray at > When you start exploring the converse of Fermat’s little theorem, you’ll discover that it’s true for a lot of numbers. For example, for any integer a, the number a^2–a is divisible by 2. You can see this by factoring a^2–a as a×(a−1). Since a and a−1 are consecutive integers, one of them has to be even, and so their product must be divisible by 2. This is Fermat's little theorem itself, NOT its converse. The latter would say that if a^p-a is divisible by p, then p is prime. To explore that, you would of course try non-prime p, as it holds vacuously for prime p. So the the author continues to "explore the converse" by trying p=3 and p=5. > In the 1800s, Bertrand’s postulate showed that for any n>3, there is always a prime number between n and 2n. Why not state the weakest possible condition of n>1 ?! > a new result about how Carmichael numbers are distributed. I was a bit surprised to see the article end here, as I was expecting to learn a little about just what this distribution is like. Fortunately, the other "story of Daniel Larsen" article that this one links to does say more on that: > In fact, Larsen’s argument didn’t just allow him to show that a Carmichael number must always appear between X and 2X. His proof works for much smaller intervals as well.
- ziyao_w 3y ago
  Good point - https://en.wikipedia.org/wiki/Bertrand%27s_postulate https://en.wikipedia.org/wiki/Bertrand%27s_postulate is actually for any n > 3 there is always a prime between n and 2n - 2, or like you said for n > 1, between n and 2n.
- tgv 3y ago
  > So the the author continues to "explore the converse" by trying p=3 and p=5. That's to set it up for the case of 561, assuming the converse holds and it is a good test for primes. It's a bit long-winded, but acceptable (IMO).
  - tromp 3y ago
    That case should be set up by trying all composites under 561 first...
LightMachine 3y ago
Why they write these articles like that? 10 pages long article about a proof... doesn't even mention what the proof is! Come on?

24 points by nsoonhui 3y ago

Wow, number theory was once my thing... Carmichael numbers are interesting indeed. Their very existence indicates that the "true nature" of primes is more obscure and complicated than people of the earlier ages thought. Some applies to regular and irregular primes. If all primes were regular, Fermat's Last Theorem would have been a lot easier to prove. It is possible that Fermat himself was mistaken and saw the outlines of the proof for regular primes. Also, maths is one of the last remaining sciences where a fresh insight from a relative newbie can still move major, century-old obstacles.

tromp 3y ago

The author appears to go astray at > When you start exploring the converse of Fermat’s little theorem, you’ll discover that it’s true for a lot of numbers. For example, for any integer a, the number a^2–a is divisible by 2. You can see this by factoring a^2–a as a×(a−1). Since a and a−1 are consecutive integers, one of them has to be even, and so their product must be divisible by 2. This is Fermat's little theorem itself, NOT its converse. The latter would say that if a^p-a is divisible by p, then p is prime. To explore that, you would of course try non-prime p, as it holds vacuously for prime p. So the the author continues to "explore the converse" by trying p=3 and p=5. > In the 1800s, Bertrand’s postulate showed that for any n>3, there is always a prime number between n and 2n. Why not state the weakest possible condition of n>1 ?! > a new result about how Carmichael numbers are distributed. I was a bit surprised to see the article end here, as I was expecting to learn a little about just what this distribution is like. Fortunately, the other "story of Daniel Larsen" article that this one links to does say more on that: > In fact, Larsen’s argument didn’t just allow him to show that a Carmichael number must always appear between X and 2X. His proof works for much smaller intervals as well.

ziyao_w 3y ago
Good point - https://en.wikipedia.org/wiki/Bertrand%27s_postulate https://en.wikipedia.org/wiki/Bertrand%27s_postulate is actually for any n > 3 there is always a prime between n and 2n - 2, or like you said for n > 1, between n and 2n.
tgv 3y ago
> So the the author continues to "explore the converse" by trying p=3 and p=5. That's to set it up for the case of 561, assuming the converse holds and it is a good test for primes. It's a bit long-winded, but acceptable (IMO).
- tromp 3y ago
  That case should be set up by trying all composites under 561 first...

LightMachine 3y ago

Why they write these articles like that? 10 pages long article about a proof... doesn't even mention what the proof is! Come on?