The question is      Prove that $a\mathbb{Z}\cap b\mathbb{Z}=[a,b]\mathbb{Z}$.   Hint:First prove $b\mid a \Leftrightarrow a\mathbb Z\le b\mathbb Z$ and then prove $a\mathbb Z+b\mathbb Z=(a,b)\mathbb Z$   I managed to prove $b\mid a\Leftrightarrow a\mathbb Z\le b\mathbb Z$. About the next claim: From the first statement ($b\mid a...$) it follows that since $(a,b)\mid a$ and $(a,b)\mid b, a\mathbb Z+b\mathbb Z\le(a,b)\mathbb Z$. My problem is proving the opposite side (a.k.a $a\mathbb Z+b\mathbb Z\ge(a,b)\mathbb Z$). How can I do so and how can I proceed from $(a,b)$ to $[a,b]$?
124150
Let be $$ a_1,\ldots,a_r \in \mathbb{N} $$  I want to show that  $$  a_1 \Bbb Z \cap \dotsb \cap a_r \mathbb{Z} = {\rm lcm}(a_1, \ldots, a_r) \Bbb Z$$  ok,I know that $ {2}\Bbb Z \cap 3\mathbb{Z} = 6\mathbb{Z}. $ I struggle getting an idea of how to prove it correctly, maybe you guys can help!
120354
I hope to show that $$\mathbb{Z}[i]/(a+bi) \cong \mathbb{Z}_{a^2+b^2}$$ for $(a,b)=1$.  I made an effort to find a homomorphism but I failed.  Can you give a hint?
111485
How to prove that if A, B, C are submodules of M module then $$ A \cap (B+C) = (A\cap B)+C \iff C \subseteq A $$  Any help would be appreciated!
16752
$A\setminus B = A \setminus B \cap A$  Proof $A\setminus B \subseteq A \setminus B \cap A$:  $x\in A\setminus B \to x\in A \land x \notin B$  $x\in A \land (x \notin B \color{red}{\lor x\notin A})$  $x \in A \land (x\notin B \cap A)$  $x \in A\setminus B\cap A$  Proof $A \setminus B \cap A \subseteq A\setminus B$:  $x \in A \setminus B \cap A \to x\in A \land x\notin B \cap A$  $x \in A \land (x\notin B \color{red}{\lor x\notin A})$  $x \in A \land x \notin B$  $x \in A \setminus B $  Therefore $A\setminus B = A \setminus B \cap A$  I am confused how the stuff in the red is formed. Can we add an or statement to anything and ignore if its true or not if the first part of the or is true?
34206
I need help with this proof:     Let $A$ and $B$ be sets. Prove that $A\cap (B\setminus A)\subseteq \varnothing$.   My problem: Since $B\setminus A$ = $B\cap A^{c}$, we can say that if $x \in B$ then $x\in B\cap A^{c}$and $x \in A^{c}$ as well. So $x\notin A$.  Where do I go from here?
74091
Problem: Show that if $(b,c)=1$, then for any integer $a$ we have $(a,bc)=(a,b)(a,c)$. [Hint: Prove that each member of the alleged equation divides the other.]  This is the way I proved it (without the hint): $(a,b)(a,c)=(a(a,b),c(a,b))=(a^2,ab,ac,bc)=(a^2,bc,a(b,c))=(a^2,a,bc)=((a^2,a),bc)=(a,bc)$  But how should I prove it using that hint or at least a similar proof.
33423
I want to show that for $\gcd(a,b) = 1$ $a,b \in Z$   $\gcd(a+b, a-b) = 1$ or $\gcd(a+b, a-b) = 2$ holds.      I think the first step should look something like this:      $d = \gcd(a+b, a-b) = \gcd(2a, a-b)$   From here I tried to proceed with two cases. 1: $a-b$ is even, which leads to $\gcd(a+b, a-b) = 2$ 2: $a-b$ is odd, which leads to $\gcd(a+b, a-b) = 1$  My main problem I think is, that I do not know how I should include $\gcd(a,b) = 1$ in the proof.  Any help is appreciated. Thx in advance. Cherio Woltan
27132
I want to prove that if $d \mid n \in \mathbb{N}$, then $\varphi(d) \mid \varphi(n)$.  It's given that $d \mid n$, so we know that $n = dm$, for some $m \in \mathbb{Z}$. Now, I want to show that $\varphi(d) \mid \varphi(n)$, i.e. that $\varphi(n) = k\varphi(d)$, for some $k \in \mathbb{Z}$. Drawing a blank of where to go from here.
16440
I was doing the problem $$ A+B=AB\implies AB=BA. $$  $AB=BA$ means they're invertible, but I can't figure out how to show that $A+B=AB$ implies invertibility.
61252
Let $a,b$ be positive integers such that $a\mid b^2 , b^2\mid a^3 , a^3\mid b^4 \ldots$ that is $a^{2n-1}\mid b^{2n} ; b^{2n}\mid a^{2n+1} , \forall n \in \mathbb Z^+$ , then is it true that $a=b$ ?
38627
Let $a$ and $b$ be integers. Show that if $(a,b) = 1$ then $(a,a + b) = 1$.   I'm stuck on how to do this proof. I tried using the fact that $ax+by=1$ but that didn't seem to help.
46007
How to prove $x^a-1\mid x^b-1 \iff a \mid b$, where $x \ge 2$ and $a,b,x \in \Bbb Z$.  I've tried the following in attempting to solve this:  $$a\mid b \Rightarrow aq=b \Rightarrow x^{aq}=x^b \Rightarrow x^ax^q=x^b$$  Because $x^q \in \Bbb Z$, it follows that $x^a\mid x^b$.  This is as far as I have gotten; any help getting further is appreciated.  Note: It may be that this identity is not true at all?
113901
Prove the following: If $a \mid bc$, then $a  \mid \gcd(a, b)c$.   I tried to set $\gcd(a, b)$ to $b$ and used the fundamental theorem of arithmetic to prove that it is divisible by $a$, but I can't prove that $a \mid bc$, if and only if $a\mid b$ and $a\mid c$. Please help. Thanks.
50999
How would I formally prove that for the integers $a$, $b$, and $d$, if $d\mid da+b$, then $d\mid b$? Would a direct proof be the best option? If I do a direct proof I seem to get stuck pretty quickly... in fact I don't even know how to start...   Assume $d\mid da+b$ ... ?  This might look like a HW dump but I honestly just don't know how to continue...
18363
I need to prove that $z\gcd(a,b)=\gcd(za,zb)$.    I tried a lot, for example, looking at set of common divisors of the two sides, but I can't conclude anything from that. Can you please give me some advice how I can handle this problem? And $a,b,z \in \mathbb{Z}$.
109275
Prove that if $a,b,c$ are integers and both $\frac{a}{b} + \frac{b}{c} + \frac{c}{a}$ and $\frac{a}{c} + \frac{b}{a} + \frac{c}{b}$ are integers, then $|a|=|b|=|c|$.   Well this is what I have done so far:  From the fact that $\frac{a}{b} + \frac{b}{c} + \frac{c}{a}$ is an integer, we get  $$abc \mid ab^2 + bc^2 + ca^2 \tag{1}$$  In the same way, we also have  $$abc \mid a^2b + b^2c + c^2a \tag{2}$$  so  $$\begin{align} abc \mid a(ab^2 + bc^2 + ca^2) - b(a^2b + b^2c + c^2a) &amp;\Rightarrow abc \mid c(a^3-b^3) \tag{3}\\ &amp;\Rightarrow ab \mid a^3-b^3 \tag{4}\\ &amp;\Rightarrow ab \mid b^3(a^3-b^3) \tag{5} \end{align}$$ and so $a \mid b^6$. In the same way, we can also get $b \mid c^6$ and $c \mid a^6$.      But what should I do after this?   Any help is surely appreciated! Thanks!
87766
Let $a$, $b$, and $c$ be integers, where a $\ne$ 0. Then $$ $$ (i) if $a$ | $b$ and $a$ | $c$, then $a$ | ($b+c$) $$ $$ (ii) if $a$ | $b$ then $a$|$bc$ for all integers $c$; $$ $$ (iii) if $a$ |$b$ and $b$|$c$, then $a$|$c$.  Prove that if $a$|$b$ and $b$|$c$ then $a$|$c$ using a column proof that has steps in the first column and the reason for the step in the second column.  My book is really vague. I'm not really sure what to do..  First I was thinking something like this: $a\mid b\Rightarrow b=as$ and  $b\mid c\Rightarrow c=bt$ and $a\mid c\Rightarrow c = au$ $$ $$ $b + c + c = b + 2c$  $$ $$ $=as + 2(bt+au)$  Yea then I just got lost.