I named $\gcd(a,m) = d$ and $\gcd(ab,m) = d' $   So I know that $d\mid a$,  $d\mid m $  and  $d'\mid ab $ , $d' \mid m$   But I can't use the transitive property of divisibility here.  How can I prove that $d \mid d'$?
22190
So far I have that $a|b$ implies $b=ax$ for some x in the integers and $c|d$ implies that $d=cy$ for some y in the integers.  From here I can see that $gcd(a,c)|gcd(b,d)$ is logically equivalent to $gcd(a,c)|gcd(ax,cy)$ but I am not sure where to go from there.
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...
41678
As stated in the title, the problem to prove is  Let $a,b,c \in \mathbb{Z}$. If $\gcd(a,c)=1=\gcd(b,c)$, then $\gcd(ab,c)=1$.  I think I've proved it, but I would like a second opinion. Here goes:  PROOF Suppose $\exists d \neq 1 \in \mathbb{Z}$ s.t $\gcd(ab,c)=d$. It follows then that $d|ab$ and $d|c$. Now, if $d|a$ but not $b$, then since $d|c$, by linearity we have $d|\gcd(a,c)$. But $\gcd(a,c)=1$ so then $d=1$. The argument for $d|b$ is similar, resulting in $d=1$. Thus, since both cases lead us to $d=1$, we have a contradiction on our hands $\implies$ $ab$ and $c$ must be relatively prime $\implies \gcd(ab,c)=1$. $Q.E.D$  The question: was this a valid argument or do you think I should provide more?
5328
Let $m \in \mathbb{Z^+} , n \in \mathbb{Z^+}$ and let $d=\gcd(m,n)$. Prove that  $$ m\mathbb{Z}+n\mathbb{Z}=d\mathbb{Z}. $$  My attempt is use inclusion to show. Let $a \in m\mathbb{Z}+n\mathbb{Z}$. Then we have $a=ms+nt$. Since $d=\gcd(m,n)$, by Bézout's Lemma, we have $d=ms+nt$ for some integers $s$, $n$. Hence, we have $a=ms+nt=d \in d\mathbb{Z}$. Let $a \in d \mathbb{Z}$. Then we have $a=db=msb+ntb \in m\mathbb{Z}+n\mathbb{Z}$. Is my proof valid?  Remark: the following is the part 2 of the question. Prove that $m\mathbb{Z}\cap n\mathbb{Z}=\frac{mn}{d}\mathbb{Z}$. This one I have no idea how to start
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 $.
61279
Does anyone have a intuition or argument or sketch proof of why $\gcd(a,b) = \gcd(b, a \pmod b)$?  I do have a proof and I understand it, so an intuition would be more helpful.    The proof that I already have:  I show $\gcd(a,b) \mid \gcd(b, a \pmod b)$ and $\gcd(b, a \pmod b) \mid \gcd(a, b)$ which implies $\gcd(a,b) = \gcd(b, a \pmod b)$ and stuff is non-negative.  WLOG $a \geq b$  $\gcd(a,b) \mid a$  $\gcd(a,b) \mid b$  so it divides any linear combination of a and b  Since $a \pmod b = a - qb$ then:  $\gcd(b, a - qb) = bx + (a-qb)y$  $\gcd(b, a - qb) = bx + ay - qby $  $\gcd(b, a \pmod b) = b(x-qy) + ay$  which is a LC of $a$ and $b$.  So $\gcd(a,b) \mid \gcd(b, a \pmod b)$.  Other direction is nearly identical.
2131
I know that $\gcd(a,b)$ divides $a$ and $b$, and must also then divide $(a)(t)$ ($t$ being some integer).  This makes sense to me, but how do I prove it?  It seems that the addition of $(a)(t)$ is a continuation of the linear combination of $\gcd(a,b) = av + bu$ for some $v$, $u$ being elements of $\mathbb{Z}$.    Any help?
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!
10327
I know that the definition of gcd of two numbers $ a, b $ is $G$ ,where  $G\mid a$, $\;G\mid b  $ and  If $d\mid a$, $\;d\mid b $, then $ d\mid G.$  Now, using this how do I prove that in $\mathbb Z$, this definition coincides with the definition that $G$ is the GREATEST (w.r.t. usual ordering) common divisor?  So to prove that I write, say $G=dq+ r,\;  0\le r&lt;d$. My aim is to show that $ r$ is $0$. So I have to construct some $G+x $ which is also a divisor of both $ a$ and $b$, to get a contradiction. How do I get $ x$?
65951
How can I find $\gcd(a^m+1,a^n+1)$ with $a,m,n$ positive integers?    I have this idea:  Let $d=\gcd(m,n)$. Then there exist positive integers $x,y$ such that $mx-ny=d$ (WLOG). We shall find $G=\gcd(a^m+1,a^n+1)$.  If $m,n$ are odd, then $d$ is odd, therefore one and only one of $x,y$ is even. We have:  $$a^{ny}(a^d+1)=a^{mx}+a^{ny}=(a^{mx}-1)+(a^{ny}+1)=(a^{mx}+1)+(a^{ny}-1).$$  If $x$ is even and $y$ is odd, then $a^{m}+1\mid a^{mx}-1$ and $a^{n}+1\mid a^{ny}+1$, therefore $G\mid a^{ny}(a^d+1)$. If $x$ is odd and $y$ is even, then $a^{m}+1\mid a^{mx}+1$ and $a^{n}+1\mid a^{ny}-1$, thus $G\mid a^{ny}(a^d+1)$. However, since $\gcd(a^m+1,a^{ny})=\gcd(a^n+1,a^{ny})=1$, so $\gcd(G,a^{ny})=1$, hence $G\mid a^d+1$. We also have $a^d+1\mid a^{m}+1$ and $a^d+1\mid a^{n}+1$, so $a^d+1\mid G$. Thus $G=a^d+1$.  If $v_2(m)=v_2(n)=v_2(d)=k&gt;1$, then there exist some odd numbers $m_1,n_1,d_1$ such that $m=2^km_1,n=2^kn_1,d=2^kd_1$. We shall have $m_1x-n_1y=d_1$, so one and only one of $x,y$ is even, and we can use the same argument when $m,n$ are odd, so $G=a^d+1$.  However, if $v_2(m) \neq v_2(n)$, I cannot find any solutions for this. I think that $G \in \{1,2\}$, but I cannot prove or disprove it. How can I find $G=\gcd(a^m+1,a^n+1)$ if $v_2(m) \neq v_2(n)$ ? Moreover, is there anything from my arguments that can be improved ?
103058
Prove that (a) gcd(a, b) = gcd (a, b – a) (b) Let r be the remainder if we divide b by a. Then gcd(a, b) = gcd(a, r).  I solved part a like: Assume a=pcommonpa b=pcommonpb gcd (a,b) = pcommon b-a = pcommon*(pb-pa)  So: gcd(a, b-a) = common again But I need help for part b guys I can solve it but I need a mathematical expression.
80346
I'm having some trouble proving the following proposition:  Let $m \in \mathbb N$ and $a,b,k \in \mathbb Z$, then: $$ka \equiv kb \pmod{m} \iff a \equiv b \pmod{\frac{m}{\gcd(k,m)}}$$  I began by trying to prove that $a \equiv b \pmod{\frac{m}{\gcd(k,m)}} \implies ka \equiv kb \pmod{m}$ but I got stuck with: $$a\gcd(k,m) \equiv b \gcd(k,m) \pmod{n}$$ Then I also tried to prove that $ka \equiv kb \pmod{m} \implies a \equiv b \pmod{\frac{m}{\gcd(k,m)}}$ But I also was not able to conclude anything. This is my first time learning modular arithmetic and I'm still confused with the concepts and have no intuition with this at all, this is perhaps why I'm having some trouble solving these types of questions. How can I prove this?
72040
Show that if $fd'=f'd $ and the pair $f, d $ are coprime, that is gcd($d,f $)=1, as is the pair $f',d' $, then $f=f' $ and $d=d' $.  This should be simple, but I couldn't verify it. Thanks in advance!
80532
I'm trying to prove that $\operatorname{lcm}(n,m) = nm/\gcd(n,m)$ I showed that both $n,m$ divides $nm/\gcd(n,m)$  but I can't prove that it is the smallest number.  Any help will be appreciated.
50039
I am having trouble completing the proof if $a \mid b$ then $\gcd(a,b) = |a|$. My attempt: If $a\mid b$ there is a $k \in \mathbb{Z}$ such that $b = ak$. Let $d = \gcd(a,b)$. Then $d \mid a$ and $d\mid b$ so by hypothesis we have $d\mid a$ and $d\mid ak$. Hence, $d \leq \gcd(a,ak)$. We can easily see that the largest common denominator of $a$ and $ak$ is $|a|$, thus $d \leq\gcd(a,ak) = |a|$ so $\gcd(a,b) \leq |a|$. I am stuck and not sure how to show the reverse inequality $|a| \leq \gcd(a,b)$ to complete the proof.
83612
Given that $\gcd(a,b) = 1$ and $\gcd(c,d) = 1$, show that $\gcd(ac,bd) = \gcd(a,d) * \gcd(b,c)$.  The work that I have done so far goes as follows.  We write  $$\gcd(ac,bd) = acv + bdu$$ then  $$\gcd(a,d) = ax + dy\quad\text{and}\quad\gcd(b,c) = bs + ct$$ and so \begin{align} \gcd(a,d) * gcd(b,c) &amp;= (ax + dy) * (bs + ct) \\ &amp;= axbs + axct + dybs + dyct \end{align} I tried to factor out some terms, and get $ax(bs+ct) + dy(bs+ct)$  but then I am stuck here. I don't know what else to use to prove the answer.
83944
I was told to prove $\gcd(ab,c)=1$ then $\gcd(a,c)=1$.  I picked a number $p$ that goes into $ab$ and $c$, so $ab=px$ and $c=py$. but now what?? I tried $abc=p^2xy$ but then I can't.  Please help me!