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k a January Highlights and 2025 AoPS Online Class Information
jlacosta   0
Jan 1, 2025
Happy New Year!!! Did you know, 2025 is the first perfect square year that any AoPS student has experienced? The last perfect square year was 1936 and the next one will be 2116! Let’s make it a perfect year all around by tackling new challenges, connecting with more problem-solvers, and staying curious!

We have some fun new things happening at AoPS in 2025 with new courses, such as self-paced Introduction to Algebra B, more coding, more physics, and, well, more!

There are a number of upcoming events, so be sure to mark your calendars for the following:

[list][*]Accelerated AIME Problem Series classes start on January 6th and 7th. These AIME classes will run three times a week throughout the month of January. With this accelerated track, you can fit three months of contest tips and training into four weeks finishing right in time for the AIME I on February 6th.
[*]Join our Math Jam on January 7th to learn about our Spring course options. We'll work through a few sample problems, discuss how the courses work, and answer your questions.
[*]RSVP for our New Year, New Challenges webinar on January 9th. We’ll discuss how you can meet your goals, useful strategies for your problem solving journey, and what classes and resources are available.
Have questions? Our Academic Success team is only an email away, write to us at success@aops.com.[/list]
AoPS Spring classes are open for enrollment. Get a jump on 2025 and enroll in our math, contest prep, coding, and science classes today! Need help finding the right plan for your goals? Check out our recommendations page!

Don’t forget: Highlight your AoPS Education on LinkedIn!
Many of you are beginning to build your education and achievements history on LinkedIn. Now, you can showcase your courses from Art of Problem Solving (AoPS) directly on your LinkedIn profile! Don't miss this opportunity to stand out and connect with fellow problem-solvers in the professional world and be sure to follow us at: https://www.linkedin.com/school/art-of-problem-solving/mycompany/ Check out our job postings, too, if you are interested in either full-time, part-time, or internship opportunities!

Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

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0 replies
jlacosta
Jan 1, 2025
0 replies
Exercise 1.1
Alohomora.   0
Dec 10, 2024
What I've tried so far:

<Describe what you have tried so far here. That way, we can do a better job helping you!>
Let $m$ be the greatest number in the finite set $\mathbb{N}_0\backslash N$. Take an arbitrary number $p$ in $N$. We have: $$p\mid (p+m) + (p+m),\  p\nmid p+m$$So $p$ is not prime and therefore $N$ doesn't have any prime.

Where I'm stuck:

<Describe what's confusing you, or what your question is here!>
0 replies
Alohomora.
Dec 10, 2024
0 replies
Exercise 0.4
Thayaden   3
N Nov 9, 2024 by felixgotti
In general, $\mathbb{Z}_n$ is a ring. A ring is and integral domain iff $0$ is the only zero-divisor. Another way of putting this is for some prime number $p$ show that,
$$ ab \not \equiv 0\pmod{p}$$where $a,b\in \mathbb{Z}_n$ where $a,b\not\equiv 0$. Consider $ab\equiv 0\pmod{p}$ for some $a,b\in\mathbb{Z}_n$ where $p$ is still some prime number. This implies $p|ab$ although by definition $p|pk$ ONLY for some $k\in \mathbb{Z}$ this implies that either $a=p$ or $b\equiv p$ although in modulus $p$ we have $p\equiv 0$ thus $a$ or $b$ must be a zero-divisor thus we can conclude that for any given prime $n$ $\mathbb{Z}_n$ is an integral domain. Consider $n=xy$ if we have $ab\equiv 0\pmod{n}$ that implies that $ab|xy$, in this case, $a=x$ or $a=y$ where $b$ is equal to the other that implies that $x$ is a zero divisor and if $n$ is non-prime that implies that for some non-$0$ divisor, $x$ is a zero divisor.
3 replies
Thayaden
Sep 18, 2024
felixgotti
Nov 9, 2024
Exercise 0.3
Thayaden   2
N Nov 7, 2024 by Thayaden
Crazy Idea but $(M,\gcd )$ I think the condtions work

$$\gcd(3,7)=\gcd(7,3)$$and
$$\gcd(3,7)=\gcd(3,101)$$yet
$$7\neq101$$
2 replies
Thayaden
Sep 16, 2024
Thayaden
Nov 7, 2024
Exercise 0.8
Thayaden   1
N Oct 31, 2024 by felixgotti
Consider $f(x)\in R[x]$ and such that $f^{-1}(x)\in R[x]$ we clearly see that,
$$f(x)\cdot f^{-1}(x)=1$$Recall $\text{deg}(1)=0$ thus $\text{deg}(f(x)\cdot f^{-1}(x))=1$ we can clearly see that as they are each other inverse that $\deg(f)=\deg(f^{-1})=1$ thus let,
\begin{align*}
f(x)&= a,\\
f^{-1}&=b.
\end{align*}Thus we see,
$$a\cdot b=1$$In other words $a$ and $b$ are elements in the group of units of $R[x]$ although since $f(x),f^{-1}(x)\in R$ it is also the group of units of $R$ therefor $R[x]^{\times}=R^{\times}$
1 reply
Thayaden
Oct 17, 2024
felixgotti
Oct 31, 2024
units...
Thayaden   1
N Oct 31, 2024 by felixgotti
Part 1:
Notice $\mathbb{Z}^{\times}=\{\pm1 \}$ thus for $\mathbb{Z}[i]$ we might enharite this too. This might be easily proven by letting the coefficient of the unreal part be $0$. Thus we might know that at least $\pm1$ is a unit. Taking that idea once again letting the real part be $0$ we have,
$$i\cdot-i=-(i^2)=1$$Thus as we are communitive $\pm i$ is in the set this makes sense from an algebraic perspective on the complex plane when multiplying a unit we have a change of direction likewise this is also true for real numbers thus $\mathbb{Z}[i]^{\times}=\{\pm1,\pm i \}$
1 reply
Thayaden
Sep 25, 2024
felixgotti
Oct 31, 2024
Exercise 0.5 crazy
Thayaden   1
N Oct 31, 2024 by felixgotti
Let $S$ be a finite integral domain that is not a field. For some $t\in S \not\ \{ 0\}$ there is no $t^{-1}\in S$ for some many $t$ as $R$ is not a field. Consider the powers of $t$,
$$t,t^2,t^3,t^4, ...$$let,
$$\underbrace{t,t^2,t^3,...,t^{|S|}}_{|S|},t^{|S|+1}$$Consider the first $|S|$ powers as unique, implying that $t^{|S|+1}$ is not unique or consider the first $|S|$ as not unique. In any given case for some $m>n$,
$$t^m=t^n$$$$t^m-t^n=0$$$$t^n \cdot (t^{m-n}-1)=0$$If $t^n=0$ that implies that $t=0$ (this can be shown inductively) and we know that $t \neq 0$ so that implies that,
$$t^{m-n}-1=0$$$$t^{m-n}=1$$$$t^{m-n-1}\cdot t=1$$Although that implies that $t$ has an inverse, that begins $t^{m-n-1}$ the only numbers that could not have an inverse indeed do have an inverse thus all numbers in $S$ have an inverse thus $S$ is a field!
1 reply
Thayaden
Sep 24, 2024
felixgotti
Oct 31, 2024
Exercise 0.7
Thayaden   0
Oct 2, 2024
If $R$ is an intergral domain then $R[x]$ is too,
Let $f(x),g(x)\in R[x]$ such that they are non zero, denote
$$f(x)=\sum_{i=0}^{n}{a_{i}x^{i}}$$$$g(x)=\sum_{j=0}^{m}{b_{j}x^{j}}$$Taking $f(x)g(x)$ let $c_{m+n}=a_{n}b_{m}$ be the leading coefficient of $f(x)g(x)$. Notice that $a_n,b_m\neq 0$ thus $c_{m+n}\neq0$ so in short for $p(x)\in R[x]$ the leading coefficient of $p(x)$ cannot be $0$ thus $R[x]$ must be an integral domain.
Going the other way say $R[x]$ is an integral domain,
For the sake of contradiction let $ab=0$ such that $a,b\in R$ and $a,b\neq0$, let $f(x)=a, g(x)=b\in R[x]$ taking there product,
$$f(x)g(x)=ab=0$$This contradicts our previous statement thus $R$ is an integral domain!
Finally iff $R$ is an integral domain then $R[x]$ must also be an integral domain $\blacksquare$
0 replies
Thayaden
Oct 2, 2024
0 replies
Exercise 0.2
Thayaden   0
Sep 16, 2024
Let $\{S_a \}_{a\in n}$ be a submonoid of $M$, where $a$ runs over $n$ is the index set. Let $S$ be the intersection of all possible submonoids of $M$. Notice now we just need to prove $S\subseteq M$. Let $e$ be the identity of $M$ notice that $e \in S_a$ and in turn $e\in S$ therfor $S$ has the identity of $M$ the first characteristic of a submonoid. Now we just need to prove that it is a monoid with respect to the operation of $M$. Let $p,q \in S$ by intersection rules $p\in S_a$ and $q \in S_a$ and notice $S_a$ is a submonoid in itself and thus the operation $p*q\in S_a$ for every $a$ thus $p*q\in S$ by intersection rules again. Thus $S$ is a submonoid in itself!

Let $\{S_a \}_{a\in n}$ be a subgroup of $M$, where $a$ runs over $n$ is the index set. Let $S$ be the intersection of all possible subgroups of $M$. Notice now we just need to prove $S\subseteq M$. Let $e$ be the identity of $M$ notice that $e \in S_a$ and in turn $e\in S$ therfor $S$ has the identity of $M$ the first characteristic of a subgroup. Now we need to prove that it is a subgroup with respect to the operation of $M$. Let $p,q \in S$ by intersection rules $p\in S_a$ and $q \in S_a$ and notice $S_a$ is a subgroup in itself and thus the operation $p*q\in S_a$ for every $a$ thus $p*q\in S$ by intersection rules again. Finally, we need to show that every element has an inverse. Let $x\in S$ and let $x^{-1}$ represent the inverse of $x$. Notice if $x\in S$ then it follows that $x\in S_a$ although notice that all $S_a$ are subgroups of $M$ and thus $x^{-1}\in S_a$ and since $x^{-1}\in S_a$ for any $a\in n$ then by intersection $x^{-1}\in S$ thus $S$ is a subgroup of $M$
0 replies
Thayaden
Sep 16, 2024
0 replies
Exercise 0.1(part 1)
Thayaden   1
N Sep 11, 2024 by Thayaden
let $b\in M$ by definition,
$$(1):~~e*b=b*e=b,$$$$(2):~~e'*b=b*e'=b$$this is true for all $b$ thus let $b=e'$ this might lead us to (from $(1)$),
$$e*e'=e'*e=e'$$$$e*e'=e'$$Now let $b=e$ (using $(2)$),
$$e'*e=e*e'=e$$$$e*e'=e$$By the transitive property,
$$\boxed{e'=e}$$as the first condition is true, $e,e'\in M$ we might then say that given all identity elements of $M$ denoted as $e_i$ we might have,
$$e_1=e_2$$and inductively,
$$e_k=e_{k+1}$$thus,
$$e_1=e_2=e_3=e_4=\cdots$$. Or in plain text, we might say that as one identity is equal to another then if we have some identity it is the same as the identity we already have.
1 reply
Thayaden
Sep 11, 2024
Thayaden
Sep 11, 2024
Exercise 3.4
FoxProdigy   1
N Aug 19, 2024 by felixgotti
Problem Statement: Prove that a monoid is a GL-monoid if and only if it is prime-like.

Proof:

Prime-like$\implies$ GL:
Let $M$ be a prime-like monoid. Suppose $a, b, c \in M$ such that $\text{gcd}(a, b) = \text{gcd}(a, c) = 1$. Suppose further towards a contradiction that $\text{gcd}(a, bc) \neq 1$; that is, that there exists some divisor $d \mid a$ with $d \mid bc$. Then because $M$ is prime-like, $d$ has some divisor $e$ with either $e \mid b$ or $e \mid c$. Either way, we also must have $e \mid a$, which means that either $\text{gcd}(a, b) \neq 1$ or $\text{gcd}(a, c) \neq 1$. This is a contradiction. Thus, $\text{gcd}(a, bc) = 1$, which means $M$ is GL.

GL$\implies$ prime-like:
Let $M$ be a GL monoid, and suppose $a, b, c \in M$ with $a \mid bc$. Assume towards a contradiction that for all $d \mid a$, $d \nmid b$ and $d \nmid c$ (in other words, $a$ shares no divisors with $b$ or $c$). This implies that $\text{gcd}(a, b) = \text{gcd}(a, c) = 1$. Since $M$ is GL, this means that $\text{gcd}(a, bc) = 1$. This contradicts the assumption that $a \mid bc$. Thus, there must exist some $d \mid a$ with $d \mid b$ or $d \mid c$, which means $M$ is prime-like.
1 reply
FoxProdigy
Aug 3, 2024
felixgotti
Aug 19, 2024
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