A new rare high-rank elliptic curve, and an orchard of Diophantine equations

197 pointsposted 6 days ago
by mathgenius
(thehighergeometer.wordpress.com)

25 Comments

fermigier

5 days ago

I broke the record back in 1992: Fermigier, Stéfane - Un exemple de courbe elliptique définie sur Q de rang ≥19. (French) [An example of an elliptic curve defined over Q with rank ≥19] C. R. Acad. Sci. Paris Sér. I Math. 315 (1992), no. 6, 719–722.

And again in 1997: Fermigier, Stéfane - Une courbe elliptique définie sur Q de rang ≥22. (French) [An elliptic curve defined over Q of rank ≥22] Acta Arith. 82 (1997), no. 4, 359–363.

Noam Elkies was already a major contributor to the field at the time. I dropped math to do computer stuff :)

anyon_fusion

5 days ago

what made you drop math for computer stuff, if you don't mind sharing? (I'm in a similar boat)

fermigier

5 days ago

I guess I thought that I could make a bigger dent in the universe as an open source entrepreneur than as a mathematician. Also, the fact that, at the time, the decision to switch course was not irreversible, given a new French law that had passed shortly before (i.e. I could go back to the university where I was already tenured if I didn't succeed in 6 years).

Xcelerate

6 days ago

This sounds very similar to the same process for Turing machines: https://www.quantamagazine.org/amateur-mathematicians-find-f...

Determining the halting behavior of each successive Turing machine generally becomes harder and harder until eventually we reach a machine with Collatz-like behavior.

The two problems are equivalent in some sense, but I wonder if there's an easy way to "port" over the work between the two projects.

yorwba

6 days ago

The equivalence of Diophantine equations and Turing machines is established by the MRDP theorem: https://en.wikipedia.org/wiki/Diophantine_set#Matiyasevich's...

For every Diophantine equation P(x, y) = 0, where x is a tuple of integer parameters and y is a tuple of unknowns, there is a Turing machine that takes the parameters x as input and halts iff there is an integer solution y satisfying the equation (this direction is easy, just have the Turing machine test all tuples of integers in some order and halt iff it found a solution)

and for every Turing machine, there is a Diophantine equation so that if x is an encoding of the Turing machine input, there is an integer solution y to the equation iff the the Turing machine halts for input x. (This direction is hard, you need to have y encode the execution history of the Turing machine somehow, and enforce the transition rules with the structure of the equation.)

Xen9

4 days ago

I believe there is no better kind of feeling in all of mathematics & all of science than that which one may get from the knowledge that an an intuitive, possibly shaky idea they themselves suggested already exists as rigorous and useful theorem.

ykonstant

5 days ago

Great write-up; I need to review how GRH is used to prove bounds on the rank of elliptic curves.

Also, the talk about polynomial parametrizations reminded me of the first Diophantine equation I solved in high school: (a^2+b^2)/(a+b) = (c^2+d^2)/(c+d). I had initially thought it had finitely many solutions, but then Nikos Tzanakis corrected me and told me I am missing many. So I toiled for two entire days and found the complete 2-parameter polynomial family of solutions.

user

6 days ago

[deleted]

miovoid

6 days ago

[deleted]

madars

6 days ago

Your E/Fp has order 2^3 * 3 * 37991 * 21183269 * 373015308871 * 16071902378831708724506232718210977087913221837027589 and thus you can't hope for more than 86 bits of security due to Pohlig–Hellman, never mind cofactor attacks. encrypt() is also insecure (xor every byte of the message with the same shared secret byte), even if you chose a better curve.

tptacek

6 days ago

This is much better version of the sibling comment but I'm a message board nerd and can't keep myself from pointing out that this code is probably a little bit tongue-in-cheek.

leijurv

6 days ago

`for char in message: encrypted_char = ord(char) ^ (shared_secret[0] % 256)`

This is not real encryption, it picks only one byte of shared secret and XORs it into the plaintext. Therefore, there are only 256 possible decryption keys to check, which is trivial.

Instead, you'd want to use the shared secret as a key to something strong and symmetric like AES.

thechao

6 days ago

Any idiot knows not to use power-of-two! You gotta use "+13", which is prime and, therefore, *secure*.

BobbyTables2

6 days ago

And Twice is nice!

Jerrrrrrry

4 days ago

and more than thrice increases your chances of collision by

tptacek

6 days ago

I don't think it's meant to be real encryption.

leijurv

6 days ago

I suspect it was, given that they've now deleted their comment.

user

6 days ago

[deleted]

fny

6 days ago

I feel like we’ve reached an era where information provenance is of paramount importance. This has always been an issue with fabricated data sets, but the ease at which anything can be fabricated—even a video—demands some new determinant of what is real and human born.