Let a,b,c $\in\mathbb{Z}$ such that a|k, b|k and gcd (a,b)=1. Prove that ab|k.  Any hints would be helpful.
3974
Using divisibility theorems, prove that if $\gcd(a,b)=1$ and $a|c$ and $b|c$, then $ab|c$.   This is pretty clear by UPF, but I'm having some trouble proving it using divisibility theorems. I was looking at , and if this could be proven, then the solution would follow immediately.
37376
Im trying to prove that, given $a,b$ with at least one of $a,b \neq 0$, $$ \gcd\left(\frac{a}{\gcd(a,b)},\frac{b}{\gcd(a,b)}\right)=1 $$ I have tried to prove the identity  $$ \gcd(c\cdot a, c\cdot b) = c\cdot \gcd(a,b) $$ with $c =  \dfrac{1}{\gcd(a,b)}$  However I'm having trouble understanding the proof.  Thank you for your time.
49507
I would appreciate hints to this. I've done part (a) but am unconfident. Wondering how I could approach part (b)    Question's comment -- The aim of this question is to use the Division Algorithm and the definition of greatest common divisor (gcd) to show that $d_0 = \gcd(a,b)$.    Question:  Let $a,b$ be integers, not both zero, let $S$ be the set of integers defined by $$S = \{ ax + by | x,y \in \mathbb{Z}\}$$ and let $d_0$ be the smallest positive integer in the set $S$. Prove: a. If $s\in S$, then $d_0$ is a divisor of $s$. b. $d_0$ is a divisor of both $a$ and $b$. c. If $d$ is a divisor of both $a$ and $b$, then $d$ is a divisor of $d_0$. d. $d_0 =\gcd(a,b),$ and hence there exists integers $x,y$ such that $ax + by = \gcd(a,b)$.    Attempt:  Note I don't expect full (or any hints to the problem(s) as there are quite a bit, but I would just appreciate hints. a. Let $s \in S$. Then $s = d_0 q + r$ for some $q,r \in \mathbb{Z}$. By definition, $d_0 = ax_0 + by_0$ for some $x_0,y_0 \in \mathbb{Z}$. $s = ax_s + by_s$ for some $x_s,y_s \in \mathbb{Z}$. Hence, $$ax_s + by_s = q(ax_0 + by_0) + r\\ a(x_s - qx_0) + b(y_s - qy_0) = r.$$ Define $x_r:= x_s - qx_0 \in \mathbb{Z}$ and $y_r:= y_s - qy_0 \in \mathbb{Z}$. Then this shows that $r \in S$ which implies $d_0 | s$.
49434
If $\gcd(a,b)=1$, $\gcd(a,y)=1$ and $\gcd(b,x)=1$ then prove that $ax+by$ is prime to $ab$.  I tried assuming Diophantine Equations for all the relations and representing everything in terms of $a$, but that didn't lead me anywhere. There must be an intuitive and elegant approach right?
24558
How to show that     $$ \gcd\bigg( {a^n-b^n \over a-b} ,a-b\bigg )=\gcd(n d^{n-1},a-b ) $$    $a,b\in \mathbb Z$   where $d=\gcd(a,b)$?    Note $\ $ Some of the answers below were merged from . The answers (and their comments) may depend on context provided in that question.
34036
Prove for positive integers $a,b,c$ and $d$  (where $b  \neq 0$ , $d \neq 0$ and $b \neq d$), if $\gcd(a,b) = \gcd(c,d) = 1$, then $\frac{a}{b} + \frac {c}{d}$ is not an integer.  I understand that if $\gcd(a,b)$ and $\gcd(c,d) = 1$, at least one number in each pair is a prime or is $1$. As for after that, I'm totally stumped, could I get some tips, clues, help?
122284
Let $a,m,n\in\mathbf{N}$. Show that if $\gcd(m,n)=1$, then $\gcd(a,mn)=\gcd(a,m)\cdot\gcd(a,n)$.  Proof:  Let $u,v\in\mathbf{Z}$ such that $\gcd(m,n)=um+vn=1$. Let $b,c\in\mathbf{Z}$ such that $\gcd(m,a)=ab+cm$. Let $d,e\in\mathbf{Z}$ such that $\gcd(a,n)=ad+en$.  So $\gcd(a,m)\cdot\gcd(a,n)=a^2bd+cmen+aben+emad$.  Where do I go from here?
52697
How can I prove that $\gcd(a, b) = \gcd (a, b + ma)$?  I have tried this:  let $g = \gcd(a, b)$, then $g \mid a$ and $g \mid b$. This means that $g \mid ax+by$.  I don't know what to do next.  Thanks.
45763
Let $p$ be a prime. Suppose that $\gcd(a, b) = p$. Find $\gcd(a^2,b)$  for all integers a and b.   I was able to prove that $\gcd(a^2,b^2) = p$ but I don't know if it helps.  Any help is greatly appreciated.
51653
Did not find this from this website...  If $$ \gcd(a,b)=1,$$ then there exists integers $x$ and $y$ such that $$xa+yb=1.$$  Now, the tip is to use particular corollary, that states:  The class $[m]_{n}$ generates $\mathbb{Z}/n\mathbb{Z}\Leftrightarrow \gcd(m,n)=1.$  I am totally lost with the corollary.  Let's assume that $\gcd(m,n)=1$. Then, $[m]_{n}$ generates $\mathbb{Z}/n\mathbb{Z}$.  OK! Then what?  There is also follow up, where I have to prove the converse. I am familiar with Bezout's lemma.
107898
Show that $\gcd(a,bc)=1$ if and only if $\gcd(a,b)=1$ and $\gcd(a,c)=1$.  I am new at proofs and I think I should use Euclid's Lemma which states "If $p$ is a prime that divides $ab$, then $p$ divides $a$ or $p$ divides $b$. However, I am not sure how to create a concrete proof or argument.  Any hints or help would be greatly appreciated. Thanks!
61758
I am trying to demonstrate that  $$((a,b),c)=(a,(b,c))\qquad\text{ (Associativity of GCD)}.$$   My proof is the following:  Since $((a,b),c)|(a,b)$, $(a,b)|b$, $(a,b)|a$ and by transitivity we have $$((a,b),c)|b, ((a,b),c)|a.\tag{1}$$ We also have $((a,b),c)|c$.   By linearity we have $\forall x,y\in \mathbb{N}$ $$((a,b),c)|bx+cy.\tag{2}$$  We can choose $x$ and $y$ so that $$bx+cy=(b,c).\tag{3}$$   Therefore $((a,b),c)|(b,c)$ and $((a,b),c)|a$.   Again by linearity $$((a,b),c)|xa+y(b,c)\tag{4}$$ and we can choose $x$ and $y$ so that $$xa+y(b,c)=(a,(b,c)).\tag{5}$$   This mean $((a,b),c)|(a,(b,c))$.   Using an analogous argument we can show that $$(a,(b,c))|((a,b),c)$$ and by the reflexivity the thesis follows.      Is this correct?
23402
(a) Prove $(k,n+k)=1$ iff $(k,n)=1$. (b) Is it true that $(k,n+k)=d$ iff $(k,n)=d$? (all variables are integers, (a,b) means gcd(a,b))  For (a), I can only get so far.  $(k,n+k)=1\ \Rightarrow\ kq_1+(n+k)q_2=k(q_1+q_2)+nq_2=1$.  Note that we don't yet have that theorem (from a different class) that says something like if $d=(a,b)$, then $d$ is the smallest (or is it biggest) number which can be made by $ak_1+bk_2$, (which I'm not quite sure how it would help, anyway). So I'm stuck.
82492
I'm learning about properties of greatest common divisors, specifically when two numbers are relatively prime.  The exercise I'm working through is :     Suppose that $\gcd(a,b) = 1$ and that $a\mid n$ and $b\mid n$. Prove that $ab\mid n$   Proof(so far):   $\gcd(a,b)=1$ implies $am+bk = 1$ for some integers $m$ and $k$.   (Given hint) Multiplying the equation by $n$ yields: $$amn+bkn=n$$  I'm not sure what to do after applying the hint. Am I supposed to show that $ab$ divides both $amn$ and $bkn$?  Thanks in advance.
1810
I know it might be too easy for you guys here. I'm practicing some problems in the textbook, but this one really drove me crazy. From $\gcd( a, b ) = 1$, I have $ax + by = 1$, where should I go from here? The extra $ac$ is so annoying. Any hint?    Thanks, Chan
45375
Show that for $ a,b ∈ Z^+,\,$ if $[a,b]=(a,b),\,$ then $a=b$.                    Prove if GCD if a and b is equal to LCM of a AND b then  $a=b$
40198
We have to prove $b|a$ and $b|c \Rightarrow b|ka+lc$ for all $k,l \in \mathbb{Z}$.  I thought it would be enough to say that $b$ can be expressed both as $b=ka$ and $b=lc$. Now we can reason that since $ka+lc=2b$ and $b|2b$, it directly follows that $b|ka+lc$?  But I'm not sure if that works for any value of $k$ and $l$ (namely $k$ and $l$ are defined through quotient between $a$ and $c$, respectively).  What am I missing?