If $\gcd(n,a)=1$, and $\gcd(n,b)=1$, prove $\gcd(n,ab)=1$
123313
If $\gcd(a,n) = 1$ and $\gcd(b,n) = 1$, then $\gcd(ab,n) = 1$. Also... $a,b$ and $n$ are natural numbers.  I feel I should begin with EEA to multiply out the gcd's, but I don't know where to go from there...
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.
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.
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?
37130
How would you go about proving that $$\gcd \left(\frac{a}{\gcd (a,b)},\frac{b}{\gcd (a,b)}\right)=1$$  for any two integers $a$ and $b$?  Intuitively it is true because when you divide $a$ and $b$ by $\gcd(a,b)$ you cancel out any common factors between them resulting in them becoming coprime. However, how would you prove this rigorously and mathematically?
43423
I tried to prove this by induction but I got stuck, this is what I've got.  Base case: $n = 2$. $\gcd(2! + 1, 3*3!+1) = gcd(3, 7) = 1.$  Inductive step: assuming for $n = k$ that $\gcd(k!+1, 3k!+1) = 1$ holds. Let $n = k + 1$. We have to prove that $\gcd((k+1)! + 1, 3(k+1)!+1) = 1.$  From assumption we get that:  $\gcd(k!+1, 3k!+1) = 1$  Multiplying this by $k + 1$ we get:  $(k + 1) * \gcd(k!+1, 3k!+1) = k + 1$  Subtracting k from both sides:  $(k + 1) * \gcd(k!+1, 3k!+1) - k = 1$  Now $k+1$ can enter the gcd and we have:  $\gcd((k +1)! + (k + 1), 3(k + 1)! + (k + 1)) - k = 1$  Now I dont know where to proceed from here.
57594
Let $R$ be a $\text{UFD}$, and let $a,b,c \in R$ such that $1=\text{gcd}(a,b).$ Suppose $a |c, \ b|c.$ Could anyone advise me on how to prove $ab |c \ ?$ How do I use the fact that every nonzero non unit $r \in R$ can be expressed as finite product of irreducible elements in $R\ ?$ Hints will suffice.
21254
How can i prove that  if $\gcd(a,b)=1$ there exist $x&gt;0$ and $y&gt;0$ such that $ax-by=1$?
38165
Suppose $\forall (a;b;c;d) \in \mathbb{Z}^4$ as $\gcd(a,b)=\gcd(c,d)=1$:  How can I prove that $\gcd(a,bc)=\gcd(a,c)$?  Also, how to prove that $\gcd(ac,bd) = (\gcd(a,d))*(\gcd(b,c))$?
9261
Hi I'm practicing some number theory and I've come across the following problem,  Suppose $n \geq 1$ is an odd integer that divides $a^2 - 1$ for some integer $a &gt; 1$. Prove that,  $$n = \gcd(a-1, n)\gcd(a+1, n)$$  Now I know that the $\gcd(a-1, a+1) = 1$ if $a$ is even and $2$ if $a$ is odd and I believe this is relevant but I can't solve the problem. I've not really any idea where to start so even just help with that would be great.  Thanks
110835
A sequence $a_{1},a_{2},\cdots $ of natural numbers satisfies   $$\gcd{(a_{i},a_{j})}=\gcd{(i,j)},\textrm{ for all } i\neq j$$      Prove that $$a_{i}=i\quad\textbf{for all } i$$   My approach is the following:  For any integer $m$,we have $$\gcd{(a_{m}.a_{2m})}=\gcd{(2m,m)}=m\Longrightarrow m \mid a_{m}$$  now I'm stuck and don't know how to proceed？
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?
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  $a=b$
2481
For all $a, m, n \in \mathbb{Z}^+$,   $$\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$$
27882
For natural numbers $a$ and $b$, show that $a \Bbb Z + b \Bbb Z = \gcd(a, b)\Bbb Z $  I just basically said that the gcd of a and b { written as C } obviously divides aZ and bZ, therefore it can be rewritten as  CdZ + CeZ -> C[dZ + eZ] -> CZ[d+e] where Cd=a and Ce=b.   if a and b and C are natural numbers, then  so are d and e. the sum of d and e is an integer, that when multiplied by an integer produces an integer. therefore  CZ[d+e] = CZ, so aZ + bZ = GCD(a,b)Z  { Where C is the gcd(a,b) }
55597
If $a,b \in\mathbb N$ and $\gcd(a,b)=1$, prove that $\gcd(a+b,a^2+b^2)$ is always equal to either 1 or 2, where $\gcd$ is the greatest common divisor. I haven't really ever solved a problem like this before, so I'd like to get some help. Thanks.
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?