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MIT PRIMES/Art of Problem Solving

CROWDMATH 2024: Generalizations of the Notion of Primes

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Generalizations of the Notion of Primes J
units...
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Thayaden
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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 $\mathbb{Z}[i]^{\times}=\{\pm1,\pm i \}$
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felixgotti
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You are right: $\pm 1$ and $\pm i$ are units. But how do you prove that these are indeed the only units?
This post has been edited 1 time. Last edited by felixgotti, Oct 31, 2024, 11:40 PM
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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
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Exercise 0.6
hkoo   1
N Jun 16, 2024 by felixgotti
What I've tried so far:

<Describe what you have tried so far here. That way, we can do a better job helping you!>

(1) Describe the group of units of the ring of Gaussian integers.

Gaussian integers is set {$a+bi \in C | a,b \in Z$}.

Case(1) a=0, b is non-zero integer
There exsits invertible (-1/b)i s.t bi*(-1/b)i = 1.
for only b=-1, $(1/b)i\in Z[i]$.

Case(2) a is non-zero integer, b=0
There exists invertible 1/a s.t a*(1/a)=1
for only a=1, $(1/a) \in Z[i]$.

Case(3) a,b are non-zero integer
To find inverse of $a+bi$, think equation $(a+bi)*(c+di)=1$.
$(a+bi)(c+di)=(ac-bd)+(ad+bc)i=1$. It means that $ac-bd=1, ad+bc=0$.
since $a,b$ are non-zero, we can know that
$c=(1+bd)/a, c=(-ad)/b$. So $(1+bd)/a = (-ad)/b$.
solve equation and we can know $(c,d)= (-b/(a^2+b^2), a/(a^2+b^2))$

since $|a|,|b| < a^2+b^2$ for non-zero integer a,b, c,d are always rational number.
Therefore we can know that {$1,-i$} is units of gaussian integers.

(2) For each n, describe the group of units of $Z_n$.
Existence of inverse of $s\in Z_n=${$0,1,...,n-1$} implies that there exist t s.t st = nQ+1.
we can solve this problem using following lemma.
$gcd(s,n)= 1$ if and only if there exists t s.t st=nQ+1 and t,Q is integer.

so group of units of $Z_n$ = {$i | (i,n)=1, 1<=i<=n$}. for example n=12, group of units of $Z_n$ ={$1,5,7,11$}. done.

proof of lemma

(1) only if
think about $i*s=n*Q_i+r_i$ for i=1,2,...,n s.t $Q_i$ is quotient, $r_i$ is residue and$0<=r_i <=n-1$. $r_i$ for i=1,..,n are different. if there exist different i,j such that $r_i=r_j$, $(j-i)*s=n*(Q_j-Q_i)$.
since (n,s)=1, n divides $(j-i)$. however n cannot divide (j-i) because $1<=|j-i|<=n-1$. so it's contradiction.
therefore {$r_i|i=1,...,n$}={$0,1,...,n-1$}. conclusivey we can know that there exits t such that residue of s*t =1.

(2) if
suppos that (s,n)=g such that gs'=s, gn'=n. since st-nQ=g(s't-n'Q)=1, g divides 1. it means that g=1. done.

Where I'm stuck:

<Describe what's confusing you, or what your question is here!>
1 reply
hkoo
May 24, 2024
felixgotti
Jun 16, 2024
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No more topics!
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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
J