The time is now - Spring classes are filling up!

MIT PRIMES/Art of Problem Solving

CROWDMATH 2024: Where Does the Goldbach Conjecture Hold?

Where Does the Goldbach Conjecture Hold? Polymath project forum
Polymath project forum
3 M G
BBookmark  VNew Topic kLocked
Where Does the Goldbach Conjecture Hold? Polymath project forum
Polymath project forum
3 M G
BBookmark  VNew Topic kLocked
G
Topic
First Poster
Last Poster
Open problem 1
drhong   8
N Nov 12, 2024 by niz3
For the case where $r$ is transcendental, we see that $\mathbb{N}[r]\equiv\mathbb{N}[x]$ so no version of goldbach’s conjecture holds. Furthermore, in the case that $r\in\mathbb{N}$, $\mathbb{N}[r]\equiv\mathbb{N}$, so the only known version is Goldbach’s weak conjecture.

Now let $r=\frac{1}{n}$ for some natural number $n$ greater than 1. Then the atoms of $\mathbb{N}[r]$ are of the form $\frac{p}{m}$ where $m$ divides a power of $n$ and $p$ is a prime not dividing $n$. Now take any $\frac{a}{n^k}\in\mathbb{N}[\frac{1}{n}]$. Then by goldbach's weak conjecture, $\frac{a}{n^k}$ can be written as the sum of at most 4 elements of the form $\frac{p}{n^k}$ for primes $p$. For each term, if $p\nmid n$, the term is an atom. Otherwise, it can be written as $\frac{1}{m}$ for some $m\mid n^k$. Let $q$ be a prime greater than $n$. By Dirichlets theorem, there is some $f$ with $1+fq$ prime. Because $1+fq>q>n$, $1+fq\nmid n$. Take some $g$ where $n^g\equiv 1 \pmod q$ and $n^g\ge 1+fq$. Then $n^g=(1+fq)+lq$ for some $l$. So $\frac{1}{m}=\frac{1+fq}{mn^g}+l\frac{q}{mn^g}$ can be wrritten as a sum of $l+1$ atoms. So every element can be written as a sum of at most $4l+4$ atoms.
8 replies
drhong
Sep 22, 2024
niz3
Nov 12, 2024
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 $S$ 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 $S[x]$" instead of $S$. I think I have the answer if it is for $S[x]$. It is enough to consider the polynomial $f(x)=ax^n+ax^{n-1}+ax^{n-2}+...+ax+a$ that cannot be expressed as the sum of a finite number of irreducibles.
4 replies
aeemc2
Nov 1, 2024
felixgotti
Nov 10, 2024
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 $S$ is an additively reduced and additively atomic semidomain, the following conditions are equivalent

(1) $\mathcal{A}(S)=S^{\times}.$

(2) every $f \in S[x^{\pm 1}]$ with |supp$(f)|>1$ can be expressed as the sum of at most 2 irreducibles.

Consider the additive monoid $M = \bigg\langle \bigg(\frac{3}{4}\bigg)^n \, | \, n \in \mathbb{N}_0 \bigg\rangle$. It is well know that $M$ is atomic and $\mathcal{A}(M)= \bigg\{\bigg(\frac{3}{4}\bigg)^n \, | \, n \in \mathbb{N}_0\bigg\} $. Since $M$ is also closed with respect to the usual multiplication of rational numbers, we know that $M$ is a semidomain. Now consider the multiplicative monoid $N= \bigg\langle \frac{3}{4} \bigg\rangle$. It is clear that $N$ is a multiplicative subset of $M$, and then we can consider the localization $N^{-1}M[x]$.

Claim 1: We see that $N^{-1}M= \bigg\langle \bigg(\frac{3}{4}\bigg)^n \, | \, n \in \mathbb{Z} \bigg\rangle$.

Proof: Let us consider an element $r\in N^{-1}M$. This means that there are $m \in M$ and $n \in N$ such that $r=\frac{m}{n}$. Take \[m=\sum_{i=1}^{k}a_i\bigg(\frac{3}{4}\bigg)^{c_i},\]for some $a_1,...,a_k, c_1,..., c_k \in \mathbb{N}_0$, and we can suppose that $n = \bigg(\frac{3}{4} \bigg)^t$ for some $t \in \mathbb{N}_0$. Note that then \[r=\sum_{i=1}^{k}a_i\bigg(\frac{3}{4}\bigg)^{c_i-t}.\]
Since $c_i-t \in \mathbb{Z}$ for every $i \in [[1,k]]$, we can conclude that $r \in \bigg\langle \bigg(\frac{3}{4}\bigg)^n \, | \, n \in \mathbb{Z} \bigg\rangle$.

Suppose now that $r\in \bigg\langle \bigg(\frac{3}{4}\bigg)^n \, | \, n \in \mathbb{Z} \bigg\rangle$. We can write then $r=\sum_{i=1}^{k}a_i\bigg(\frac{3}{4}\bigg)^{z_i}$ for some $a_1,...,a_k \in \mathbb{N}_0$, and some $z_1,...,z_k \in \mathbb{Z}$. For each $z_i$ we can find some natural numbers $z'_i$ and $z''_i$ such that $z_i=z'_i-z''_i$. Then we can write \[r=\sum_{i=1}^{k}a_i \frac{\bigg(\frac{3}{4}\bigg)^{z'_i}}{\bigg(\frac{3}{4}\bigg)^{z''_i}},\]which is clearly an element of $N^{-1}M$.
Let us denote $S=N^{-1}M$. In the paper "Puiseux monoids and transfer isomorphism", Dr. Felix Gotti proved in Proposition 3.5 that if $r\in \mathbb{Q}_{>0}$, then the additive monoid $M=\langle r^n  |  n \in \mathbb{Z} \rangle$ is atomic with $\mathcal{A}(M)=\{r^n | n \in \mathbb{Z}\} $. Then, by the theorem mentioned above, we know that the semidomain $S[x]$ satisfies Goldbach conjecture. However $M[x]$ clearly does not satisfy it. Indeed, for every $n$ the polynomial $f=\sum_{i=0}^{n}\frac{3}{4}x^{k_i}$ cannot be expressed as the sum of $k$ or fewer irreducibles.
2 replies
aeemc2
Nov 3, 2024
aeemc2
Nov 9, 2024
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 $\mathbb{N}_0[x]$ is Eisenstein's irreducible criterion that states
Criterion 1 (Eisenstein's irreducible criterion).
Let $p(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0 \in \mathbb{N}_0[x] $. If there exists a prime number $p$ satisfying the following three conditions:

- $p$ divides each $a_i$ for $0 \leq i < n$.
-$p$ does not divide $a_n$, and
-$p^2$ does not divide $a_0$.

Then the polynomial $p(x)$ is irreducible.

We have another criterion, which is even easier to check.

Criterion 2: Let $p(x) \in  \mathbb{N}_0[x]$. Suppose that $p(0) \neq 0$ and $p(1)$ is a prime number, then $p(x)$ is irreducible.

Proof
Indeed, suppose that $p(x)=q(x)r(x)$ for some $q(x), r(x) \in    \mathbb{N}_0[x]$. Note that $p(1)=q(1)r(1)$, and since $p(1)$ is prime, we know that either $q(1)=1$ or $r(1)=1$. WLOG, assume that $r(1)=1$, then $r(x)=x^n$ for some $n \in \mathbb{N}_0$. Note that if $n > 0$, then $r(0)=0$, and then $p(0)=0$, which contradicts the hypothesis. Therefore $r(x)=1$, and this implies that $p(x)$ is irreducible.

2 replies
aeemc2
Nov 1, 2024
aeemc2
Nov 3, 2024
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
0 replies
felixgotti
Nov 1, 2024
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
2 replies
felixgotti
Sep 1, 2024
felixgotti
Oct 15, 2024
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?
1 reply
BadAtMath23
Sep 23, 2024
felixgotti
Oct 1, 2024
Exercise 0.2
aeemc2   4
N Oct 1, 2024 by felixgotti
First, suppose that $n=1$. Note that we can find a polynomial of $\mathbb{N}_0[x]$ that is not irreducible. We can consider, for example, $f(x)=x^3+5x=x(x^2+5)$. Fix $n \in \mathbb{N}_{>1}$ and consider the polynomial $p(x)=x^{n}+1$. Suppose towards a contradiction that $p(x)=\sum_{i=1}^n a_i(x)$ with $a_i(x) \in \mathcal{A}(\mathbb{N}_0[x])$. Note that we have at least one $a_i(x)$ such that $ord(a_i(x)) \neq 0$. Indeed, if $ord(a_i(x))=0$ for every $i \in [[1,n]]$, then the constant coefficient of $p(x)$ is at least $n$. However if $ord(a_i(x) \neq 0$, then we have that $x \mid_{\mathbb{N}_0[x]} a_i(x)$, which is a contradiction with the irreducibility of $a_i(x)$. This proves the result.
4 replies
aeemc2
Sep 7, 2024
felixgotti
Oct 1, 2024
Exercise 0.1
Leia.math   1
N Oct 1, 2024 by felixgotti
$\left. a\right)  \Rightarrow \left. b\right) $

Let $S$ be a semiring, the Grothendieck group $\mathcal{G}\left(S\right)$ of $\left(S, +\right)$ is to be constructed by introducing inverse elements to all elements of $S$. Elements of $\mathcal{G}\left( S\right)$ are of form $m-n$ with $m$ and $n$ in $S$.

The multiplication of $S$ is extended to $\mathcal{G}(S)$ as follows:
$\left(m - n\right)\left(r - s\right) = \left(mr + ns\right) - \left(nr + ms\right),$ which is well define since $\left( mr + ns\right)$ and $\left(nr + ms\right)$ are in $S$.

As a hypothesis, $S$ is a semidomain; therefore, $S$ is a subsemiring of an integral domain $R$. Then, the monoid $\left(S, + \right)$ is contained in the group $\left(R, +\right)$. However, $\left(\mathcal{G}\left( S\right), +\right)$ is the smallest group that contains $\left(S, + \right)$, which means that $\left(\mathcal{G}\left( S\right), +\right)$ is contained in $\left(R, +\right)$.

According to this, $\left(\mathcal{G}\left( S\right), +, \cdot\right)$ is a subring of $\left(R, +, \cdot\right)$ and therefore $\left(\mathcal{G}\left( S\right), +, \cdot\right)$ is an integral domain.

$\left. b\right)  \Rightarrow \left. a\right) $

For the semiring $S$, the multiplication of $S$ extends to $\mathcal{G}(S)$ turning $\mathcal{G}(S)$ into an integral domain. But $S$ is contained in $\mathcal{G}(S)$, so $S$ is a subsemiring of $\mathcal{G}(S)$. Thus, $S$ is a subsemiring of an integral domain. Consequently $S$ is a semidomain.
1 reply
Leia.math
Sep 16, 2024
felixgotti
Oct 1, 2024
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
0 replies
felixgotti
Sep 14, 2024
0 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?
1 reply
BadAtMath23
Sep 23, 2024
felixgotti
Oct 1, 2024
a