MIT PRIMES/Art of Problem Solving
CROWDMATH 2024: Where Does the Goldbach Conjecture Hold?
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Open problem 1
drhong 8
N
Nov 12, 2024
by niz3
For the case where is transcendental, we see that so no version of goldbach’s conjecture holds. Furthermore, in the case that , , so the only known version is Goldbach’s weak conjecture.
Now let for some natural number greater than 1. Then the atoms of are of the form where divides a power of and is a prime not dividing . Now take any . Then by goldbach's weak conjecture, can be written as the sum of at most 4 elements of the form for primes . For each term, if , the term is an atom. Otherwise, it can be written as for some . Let be a prime greater than . By Dirichlets theorem, there is some with prime. Because , . Take some where and . Then for some . So can be wrritten as a sum of atoms. So every element can be written as a sum of at most atoms.
Now let for some natural number greater than 1. Then the atoms of are of the form where divides a power of and is a prime not dividing . Now take any . Then by goldbach's weak conjecture, can be written as the sum of at most 4 elements of the form for primes . For each term, if , the term is an atom. Otherwise, it can be written as for some . Let be a prime greater than . By Dirichlets theorem, there is some with prime. Because , . Take some where and . Then for some . So can be wrritten as a sum of atoms. So every element can be written as a sum of at most atoms.
8 replies
exercise 1.2
aeemc2 4
N
Nov 10, 2024
by felixgotti
Hi professor. I wanted to ask you if the correct statement for exercise 1.2 is "Let be an additively reduced semidomain containing an additive atom that is not a multiplicative unit. Prove that no analog of the Goldbach conjecture holds for " instead of . I think I have the answer if it is for . It is enough to consider the polynomial that cannot be expressed as the sum of a finite number of irreducibles.
4 replies
Problem 2
aeemc2 2
N
Nov 9, 2024
by aeemc2
This is my solution of Problem 2, part (a). Please let me know if there are any mistakes.
It was proven in Theorem 3.6 of the paper "A Golbach conjecture for Laurent Series Semidomain" that if is an additively reduced and additively atomic semidomain, the following conditions are equivalent
(1)
(2) every with |supp can be expressed as the sum of at most 2 irreducibles.
Consider the additive monoid . It is well know that is atomic and . Since is also closed with respect to the usual multiplication of rational numbers, we know that is a semidomain. Now consider the multiplicative monoid . It is clear that is a multiplicative subset of , and then we can consider the localization .
Claim 1: We see that .
Proof: Let us consider an element . This means that there are and such that . Take for some , and we can suppose that for some . Note that then
Since for every , we can conclude that .
Suppose now that . We can write then for some , and some . For each we can find some natural numbers and such that . Then we can write which is clearly an element of .
Let us denote . In the paper "Puiseux monoids and transfer isomorphism", Dr. Felix Gotti proved in Proposition 3.5 that if , then the additive monoid is atomic with . Then, by the theorem mentioned above, we know that the semidomain satisfies Goldbach conjecture. However clearly does not satisfy it. Indeed, for every the polynomial cannot be expressed as the sum of or fewer irreducibles.
It was proven in Theorem 3.6 of the paper "A Golbach conjecture for Laurent Series Semidomain" that if is an additively reduced and additively atomic semidomain, the following conditions are equivalent
(1)
(2) every with |supp can be expressed as the sum of at most 2 irreducibles.
Consider the additive monoid . It is well know that is atomic and . Since is also closed with respect to the usual multiplication of rational numbers, we know that is a semidomain. Now consider the multiplicative monoid . It is clear that is a multiplicative subset of , and then we can consider the localization .
Claim 1: We see that .
Proof: Let us consider an element . This means that there are and such that . Take for some , and we can suppose that for some . Note that then
Since for every , we can conclude that .
Suppose now that . We can write then for some , and some . For each we can find some natural numbers and such that . Then we can write which is clearly an element of .
Let us denote . In the paper "Puiseux monoids and transfer isomorphism", Dr. Felix Gotti proved in Proposition 3.5 that if , then the additive monoid is atomic with . Then, by the theorem mentioned above, we know that the semidomain satisfies Goldbach conjecture. However clearly does not satisfy it. Indeed, for every the polynomial cannot be expressed as the sum of or fewer irreducibles.
2 replies
exercise 1.3
aeemc2 2
N
Nov 3, 2024
by aeemc2
One of the irreducible criteria is well-known for polynomials with integers coefficients and also holds naturally for polynomials of is Eisenstein's irreducible criterion that states
Criterion 1 (Eisenstein's irreducible criterion).
Let . If there exists a prime number satisfying the following three conditions:
- divides each for .
- does not divide , and
- does not divide .
Then the polynomial is irreducible.
We have another criterion, which is even easier to check.
Criterion 2: Let . Suppose that and is a prime number, then is irreducible.
Proof
Indeed, suppose that for some . Note that , and since is prime, we know that either or . WLOG, assume that , then for some . Note that if , then , and then , which contradicts the hypothesis. Therefore , and this implies that is irreducible.
Criterion 1 (Eisenstein's irreducible criterion).
Let . If there exists a prime number satisfying the following three conditions:
- divides each for .
- does not divide , and
- does not divide .
Then the polynomial is irreducible.
We have another criterion, which is even easier to check.
Criterion 2: Let . Suppose that and is a prime number, then is irreducible.
Proof
Indeed, suppose that for some . Note that , and since is prime, we know that either or . WLOG, assume that , then for some . Note that if , then , and then , which contradicts the hypothesis. Therefore , and this implies that is irreducible.
2 replies
Resource 2 and Open Problem 2 posted!
felixgotti 0
Nov 1, 2024
Hi everyone!
We are happy to share that Resource 2 and Open Problem 2 (which, as Open Problem 1, has two parts) have just been posted. We hope you enjoy learning and practicing the new material as well as thinking about how to solve the new open problems.
Don't be hesitant to bring your questions and ideas to our message board!
Happy research!
Best,
Felix and Harold
We are happy to share that Resource 2 and Open Problem 2 (which, as Open Problem 1, has two parts) have just been posted. We hope you enjoy learning and practicing the new material as well as thinking about how to solve the new open problems.
Don't be hesitant to bring your questions and ideas to our message board!
Happy research!
Best,
Felix and Harold
0 replies
A New CrowdMath Project (with Resource 0 and Some Exercises)
felixgotti 2
N
Oct 15, 2024
by felixgotti
Hi everyone!
A New Research Project for CrowdMath 2024 has been released! This new research project will be about semidomains satisfying the statement of the Goldbach Conjecture.
We have the great pleasure to have Dr. Harold Polo with us, providing his direct help with this project, which is in turn motivated by the current research carried out by Dr. Polo in the intersection of semidomains and the statement of the Goldbach Conjecture.
Resource 0 (with some initial exercises) has just been posted. As always, I hope you enjoy learning the new material and working on the exercises. We plan to post the first open problem by next week.
Don't hesitate to bring your questions, solutions, and ideas to the message board!
Best,
Felix
A New Research Project for CrowdMath 2024 has been released! This new research project will be about semidomains satisfying the statement of the Goldbach Conjecture.
We have the great pleasure to have Dr. Harold Polo with us, providing his direct help with this project, which is in turn motivated by the current research carried out by Dr. Polo in the intersection of semidomains and the statement of the Goldbach Conjecture.
Resource 0 (with some initial exercises) has just been posted. As always, I hope you enjoy learning the new material and working on the exercises. We plan to post the first open problem by next week.
Don't hesitate to bring your questions, solutions, and ideas to the message board!
Best,
Felix
2 replies
Some general questions about crowd math
BadAtMath23 1
N
Oct 1, 2024
by felixgotti
Sorry if this isn’t properly suited to this form, but I would like to ask:
I was wondering roughly how much background would be good to have before trying to contribute? I am taking a course covering mathematical analysis (including some basic topology) at a local university, and group/field theory through AOPS. Would this suffice?
Also, as a bit of a more random question, why do some years the write up/publication uses just the pseudo name “CrowdMath” and other years people’s names (I assume contributors) are listed?
I was wondering roughly how much background would be good to have before trying to contribute? I am taking a course covering mathematical analysis (including some basic topology) at a local university, and group/field theory through AOPS. Would this suffice?
Also, as a bit of a more random question, why do some years the write up/publication uses just the pseudo name “CrowdMath” and other years people’s names (I assume contributors) are listed?
1 reply
Exercise 0.2
aeemc2 4
N
Oct 1, 2024
by felixgotti
First, suppose that . Note that we can find a polynomial of that is not irreducible. We can consider, for example, . Fix and consider the polynomial . Suppose towards a contradiction that with . Note that we have at least one such that . Indeed, if for every , then the constant coefficient of is at least . However if , then we have that , which is a contradiction with the irreducibility of . This proves the result.
4 replies
Exercise 0.1
Leia.math 1
N
Oct 1, 2024
by felixgotti
Let be a semiring, the Grothendieck group of is to be constructed by introducing inverse elements to all elements of . Elements of are of form with and in .
The multiplication of is extended to as follows:
which is well define since and are in .
As a hypothesis, is a semidomain; therefore, is a subsemiring of an integral domain . Then, the monoid is contained in the group . However, is the smallest group that contains , which means that is contained in .
According to this, is a subring of and therefore is an integral domain.
For the semiring , the multiplication of extends to turning into an integral domain. But is contained in , so is a subsemiring of . Thus, is a subsemiring of an integral domain. Consequently is a semidomain.
1 reply
Resource 1 and Open Problem 1 have just been posted!
felixgotti 0
Sep 14, 2024
Hi everyone!
Resource 1 and Open Problem 1 (two parts) have just been posted. I hope you enjoy learning and practicing the new material as well as thinking about how to solve the new related open problem.
Don't hesitate to bring your questions, solutions, and ideas to the message board!
Happy research!
Best,
Felix and Harold
Resource 1 and Open Problem 1 (two parts) have just been posted. I hope you enjoy learning and practicing the new material as well as thinking about how to solve the new related open problem.
Don't hesitate to bring your questions, solutions, and ideas to the message board!
Happy research!
Best,
Felix and Harold
0 replies
A New CrowdMath Project (with Resource 0 and Some Exercises)
felixgotti 2
N
Oct 15, 2024
by felixgotti
Hi everyone!
A New Research Project for CrowdMath 2024 has been released! This new research project will be about semidomains satisfying the statement of the Goldbach Conjecture.
We have the great pleasure to have Dr. Harold Polo with us, providing his direct help with this project, which is in turn motivated by the current research carried out by Dr. Polo in the intersection of semidomains and the statement of the Goldbach Conjecture.
Resource 0 (with some initial exercises) has just been posted. As always, I hope you enjoy learning the new material and working on the exercises. We plan to post the first open problem by next week.
Don't hesitate to bring your questions, solutions, and ideas to the message board!
Best,
Felix
A New Research Project for CrowdMath 2024 has been released! This new research project will be about semidomains satisfying the statement of the Goldbach Conjecture.
We have the great pleasure to have Dr. Harold Polo with us, providing his direct help with this project, which is in turn motivated by the current research carried out by Dr. Polo in the intersection of semidomains and the statement of the Goldbach Conjecture.
Resource 0 (with some initial exercises) has just been posted. As always, I hope you enjoy learning the new material and working on the exercises. We plan to post the first open problem by next week.
Don't hesitate to bring your questions, solutions, and ideas to the message board!
Best,
Felix
2 replies