Can anyone tell me if my attempt is correct?   gcd (a, a + b) = gcd (b, a + b) = gcd (a, b) = 1.  ∴, gcd (a + b, ab) = 1.
121068
I'm trying to show that if $\operatorname{gcd}(a,b) = 1$, then $\operatorname{gcd}(ab,a+b)=1$.  I've tried  to use the gcd properties: $$\operatorname{gcd}(a,b)=1 \implies  \operatorname{gcd}(a,a+b)=1\\\operatorname{gcd}(a,b)\implies\operatorname{gcd}(ab,b^2)=b$$ but I got stuck. Any hint will help.
39116
Somewhere on Stack Exchange I saw the equation     $$\gcd(2^m-1,2^n-1)=2^{\gcd(m,n)}-1.$$   I had never seen this before, so I started trying to prove it. Without success...  Can anyone explain me (so actually prove) why this equation is true?  And can we say the same when replacing the '$2$' by any integer number '$a$'?
70148
Let $D$ be a , and let $a,b,x \in D \setminus \{0\}$. Then is it true that $\gcd (ax,bx)=x \cdot \gcd (a,b)$ ?    Let $c=\gcd (a,b)$ and $d=\gcd(ax,bx)$, then as $cx|ax$ and $cx|bx$ so $cx|d$. We would be done if we could show $d|cx$, but I am unable to show that. Please help me to solve this problem.
79985
I have tried $\gcd(0,8)$ in a lot of online  gcd (or hcf) calculators, but some say $\gcd(0,8)=0$, some other gives $\gcd(0,8)=8$ and some others give $\gcd(0,8)=1$. So really which one of these is correct and why there are different conventions?
37177
Let $a,b,n$ be integers such that  $\gcd(a,n)=\gcd(b,n)=1$. How to show that $\gcd(ab,n)=1$?   In other words, how to show that if two integers $a$ and $b$ each have no non-trivial common divisor with and integer $n$, then their product does no have a non-trivial common divisor with $n$ either.     This is a problem that is an exercise in my course.   Intuitively it seems plausible and it is easy to check in specific cases but how to  give an actual proof is not obvious.
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.
19995
Here is what I got so far:  Let $\gcd(ka,km) = d$  Then by the Euclidean Algorithm, we have integers $s,t$ such that:  $$s(ka) + t(km) = d \implies k|d  $$  Let $$d/k = g = sa + tm.  \tag{1}$$  So now I need to show that $g$ is indeed = $\gcd(a,m)$. Let  $\gcd(a,m) =  h$  Now since $k|d \implies d|a , d|m  \space $  (since $$\gcd(ka,km) = d \implies  g|  \gcd(a,m) = h \\\implies h = dj  ,$$ for some integer $j$.  Putting this into (1):  $$ h/jk = g = sa + tm $$  since $$h = dj \implies dj/j = kg \\ \implies d = kg. $$   So $g$ is the $\gcd(a,m)$?  Is this a correct proof? Or have I gone in some sort of circle? Many thanks for any hints or direction!
14113
Let $a, b, c$ be integers. Prove that if $\gcd(a,b)=1$ then $\gcd(ab,c) = \gcd(a,c) \gcd(b,c)$  First time asking here. I'm not sure what your policies are on general homework help but I truly am stuck.  So far I have shown $\gcd(a,c) \gcd(b,c)$ as an integer combination of $ab$ and $c$. So if I can show that $\gcd(a,c) \gcd(b,c)$ divides $ab$ and $c$ I can use the proof that if an integer $d$ is a common divisor of $a$ and $b$, and $d=ax+by$ for some $x$ and $y$, that $d=\gcd(a,b)$. However I don't really know where to start with this. Any help would be appreciated.
69274
If  $a,b$ are positive integers such that $\gcd(a,b)=1$, then show that $\gcd(a+b, a-b)=1$ or $2$ and $\gcd(a^2+b^2, a^2-b^2)=1$ or $2 $. Progress We have   $\gcd(a,b)=1\implies \exists u,v\in \mathbb Z$ such that $au+bv=1\implies a^2u'+bv'=1, u',v'\in \mathbb Z$. Let $\gcd(a+b, a-b)=d$. Then $\mid (a+b)x+(a-b)y, \forall x,y\in \mathbb Z$. How to show $\gcd(a+b, a-b)=1$ or $2$ and $\gcd(a^2+b^2, a^2-b^2)=1$ or $2 $.
8732
If $\gcd(a,b) = 1$ and $a,b\mid x$ then $ab\mid x$.  My attempt at answering the question:  \begin{align*} x &amp;\equiv 0 \pmod{a}\\\ &amp;\Longrightarrow x\text{ is divisible by $a$}\\\ &amp;\Longrightarrow x = ma\text{ for some integer $m$}\\\ \ \\\ x &amp;\equiv 0 \pmod{b}\\\ &amp;\Longrightarrow  x\text{ is divisible by $b$}\\\ &amp;\Longrightarrow x = mb\text{ for some integer $m$}\\\ \ \\\ x^2 &amp;= (ma)(mb)\\\ x^2 &amp;= (m^2)(ab)\\\ x &amp;= \sqrt{m^2ab}\\\ x &amp;= m\sqrt{a}\sqrt{b} \end{align*} Let $m$ be $k\sqrt{a}\sqrt{b}$. Then \begin{align*} x &amp;= kab\\\ &amp;\Longrightarrow x \equiv 0 \pmod{ab} \end{align*}  Is this correct, if not can someone point me in the right direction?
4816
Could someone please help me with this proof?     Suppose that $a, b \in N$, and $d = \gcd(a, b)$. Since $d$ divides $a$, we have $a = de$ for some integer $e,$ and similarly $b = df$ for some integer $f$. Prove that $\gcd(e, f) = 1$.   I understand why it works. Since d is all the common factors of $a$ and $b, e$ and $f$ had no common factors, therefore the $\gcd(e,f) = 1$. But how do I prove this?  Thanks in advance.
110279
If integers $a$ and $b$ are not both zero, then $\gcd(a,b)=\gcd(a-b,b)$   Assume either $a=0$ or $b=0$. If $a=0$, then $\gcd(a,b)=\gcd(0,b)=b$ and $\gcd(a-b,b)=\gcd(-b,b)=-b$, so $\gcd(a,b)\neq\gcd(a-b,b)$.  If $b=0$, then $\gcd(a,b)=\gcd(a,0)=0$ and $\gcd(-b,b)=\gcd(a,0)=0$.  So, what can I say? Can you help? Can you check my proof-trying?
114208
Let $a, b, c \in \mathbb{N}$. Show that if $a^2+b^2=c^2$ then $gcd(a,b)=gcd(a,c)=gcd(b,c)$.  Using basic divisibility criteria I couldn't advance anything.
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.
66409
Let's consider two number $a$ and $b$ such as $gcd(a,b)=1$. Can you explain me intuitively why there exist $n&gt;0$ such as $a^{n} \equiv 1 \space (mod \space b)$  ?
29309
If $\gcd(a,b)=1$, then $\gcd(a^n,b^n)=1$   This seems clear, but I don't know how to prove this..  I was trying to show this by induction such that if $a^{n+1}$ = $rs$ and $b^{n+1}$ = $rt$, then $s,t$ are divisible by $a,b$ respectively, but i think this is a wrong way..
23362
Prove $\gcd(a,b,c)=\gcd(\gcd(a,b),c)$ for $0\ne a,b,c\in \Bbb{Z}$. I tried solving it with sets but I sense there are some details I am missing. I would truly appreciate your reference.