everyone  I'm stuck on this proof. I know that $\gcd(a,bc)=\gcd(a,b)\cdot \gcd(a,c)$ but I don't know how I use $\gcd(b,c)=1$ to get that $\gcd(a,bc)=\gcd(a,b)\cdot \gcd(a,c)$
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?
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 $d_0 | s$.
60939
I'm stuck at proving:     Let $a, b, c \in \mathbb{Z}, a\cdot b \neq 0$. Prove:   $$ a \mid m \hspace{15px}\text{and}\hspace{15px} b \mid m \implies \frac{a \cdot b}{\text{gcd}(a, b)} \mid m $$   My current attempt looks like this:  Let $t := \text{gcd}(a, b)$ donate the greatest common divisor of $a$ and $b$.   $$ \implies\exists\, a', b': a = t\cdot a' \hspace{15px}\text{and}\hspace{15px} b  = t \cdot b' $$  $$ \implies \frac{a  b}{\text{gcd}(a, b)} = \frac{a  b}{t} = \frac{t  a'b}{t} = a' b $$  I know that since b \mid m: $a'b \mid m$ iff $\text{gcd}(a', b) = 1$ and $a' \mid m$, but I'm not sure whether this is the right approach to this problem.   Currently I don't know about identities involving the least common multiple.
15659
Prove that if $d$ is a common divisor of two integers $a$ and $b$, then $d=\gcd(a,b)$ if and only if $\gcd(a/d,b/d)=1$.   So far I used what was given so I have $a=dk$, $b=ld$ and $\gcd(a,b)=d$ can be written as a linear combination of $ax+by=d$ but I am unsure how to use the information.   Where do I go from here? Can someone show me how to solve this using Bezout's Identity if possible?
66002
How would you show that if $ac≡bc$ $\mod m$ and $\gcd(c,m)=d$, then  $a≡b$ $\mod \frac{m}{d}$?  Any help would be much 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 $.
4209
How to prove $\gcd(a^2, b^2) = (\gcd(a, b))^2$?  My attempt:  Let $\gcd(a, b) = d$.  Then $d|a$ and $d|b$ then $d^2|a^2$ and $d^2|b^2$. i.e $d^2$ divides $a^2 ~~\&amp;~~ b^2$.
10530
Well I thought this is obvious. since $\gcd(a,b)=1$,  then we have that $a\not\mid b$ AND $a\mid bc$. This implies that $a$ divides $c$. But apparently this is wrong. Help explain why this way is wrong please.   The question tells you give me two relatively prime numbers $a$ and $b$ such that $a$ divides the product $bc$, prove that $a$ divides $c$. how is this not obvious? Explain to me how my "proof" is not correct.
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.
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.
96285
Can anyone show some hints to this? If $\gcd(a,b)=1$, then $a^{\phi(b)}+b^{\phi(a)}=1 \pmod{ab}$. I know that $a^{\phi(b)}=1 \pmod{b}$, and similarly, $b^{\phi(a)}=1 \pmod{a}$, but then how do I combine the work so that I get the result I need? Thanks!
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!
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.
85782
I guess the canonical proof to this problem is attained using the Euclidean Algorithm (I've seen some posts like this already). I came with this proof, based on gcd definition and some divisibility properties. I'd like to know if it seems correct to you.  Proof:  Let $d$ be a common divisor of both terms. Then $n! +1$ = $d$.$s$ and $(n+1)! +1 = d.t$, for $s, t$ in $\Bbb Z/{0}$. Therefore $(n +1)n! + 1 = n.n! + n! + 1 = n.n! + d.s = d.t$, and it follows that $d|n.n!$. Now, since $n=d(t-s)/n!$, $d|n$. Then $d$ must divide 1 ($n! + 1 = d.s$), and $|d|$ = 1. Being $d'$ another common divisor, it is clear that $d'$ must divide 1 also, and therefore it does divide $d$, so we have $d = gcd(n! + 1, (n+1)! +1) = 1$
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?
27965
Let $a$ and $b$ be two integers. Show that $\gcd(a^2, b^2) = \gcd(a,b)^2$.   This is what I have done so far:  Let $d = \gcd(a,b)$. Then $d=ax+by$ for some $x,y$. Then $d^2 =(ax+by)^2 = a^2x^2 + 2axby+b^2y^2$.  I am trying to create a linear combination of $a^2$ and $b^2$ but do not know what to do with the middle term.  EDIT: I would be interested in seeing a proof using the Fundamental Theorem of Arithmetic as well. I simply thought it would be easiest to use Bezout's Identity.
48672
Suppose $a, b \in \mathbb{Z}$, $a &gt; b$, and $\gcd (a,b) = 1$.  What can be said about $\gcd (a+b,a-b)$?  Is it true in general that $\gcd (a+b,a-b) \leq 2$?