Let $gcd(a, b) = d$ Let $a = rd$ and $b = sd$, where $gcd(r, s) = 1$. $\therefore gcd(r^n, s^n) = 1$ Now, how do I show that $r = 1$?
42349
I would like to show that $a^n \mid b^n$ implies $a \mid b$  I thought I could convert it to congruences and work backwards, but as far as I remember, $a \equiv b \pmod{m}$ implies $a^n \equiv b^n \pmod{m}$, not the opposite, unless $m$ is prime. Is that right?  So I am not sure how to approach this one. Any ideas?  Thanks!
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?
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.
45033
I don't really know how to prove this, i suppose it must have something to do with the definition of greatest common divisor and Bézout's identity.  I looked at some other questions here and thought that i had to get to gcd(ma+nb,mn)=1 and that gcd(a,n)gcd(b,m)=1 too, but i'm not sure.
21254
How can i prove that  if $\gcd(a,b)=1$ there exist $x&gt;0$ and $y&gt;0$ such that $ax-by=1$?
120786
If $\gcd(m,n)=1$, and $\gcd(a,m)=\gcd(b,n)=1$, then $\gcd(ab, mn)=1$.  I tried going about it algebraicallly using the definition of GCD, which gave me $$\left(a u_2+m v_2\right) \left(b u_3+n v_3\right) \left(m u_1+n v_1\right)=1$$ which, when expanded, gives $$a b m u_1 u_2 u_3+a b n u_2 u_3 v_1+a m n u_1 u_2 v_3+a n^2 u_2 v_1 v_3+b m^2 u_1 u_3 v_2+b m n u_3 v_1 v_2+m^2 n u_1 v_2 v_3+m n^2 v_1 v_2 v_3.$$ Unfortunately, factoring with respect to $mn$ and $ab$ does not give me an equation of the form $mnx+aby=1$, as I had hoped.  Then I tried going another route: Assume that $\gcd(ab,mn)=d\neq1$. Then there is a prime $p$. which divides $uab+vmn=d$. Since $p|d$ and $d|mn$ and $d|ab$, $p|mn$ and $p|ab$. Since $p$ is prime, it either divides $m$ or it divides $n$, but since $\gcd(m,n)=1$, it cannot divide both. Without loss of generality, say $p|m$.  Since $\gcd(m,a)=1$, $p\not|a$. This means that $p|b$. But I don't know how to get a contradiction from here.
108892
I am having trouble understanding why the Euclidean algorithm for finding the GCD of two numbers always works?   I found some resources here (), and here().  But I am a little confused about how they approach it here. I understand that if we have two numbers, a and b, then the greatest common divisor of a and b has to be less than a, and if a divides b, then a will have to be the GCD.   But I am confused about what happens when:  b=a*q+r So now, we are saying that we take a/r, correct? Why should we do this at all?
72815
Assume $x/a$ and $y/b$ are reduced, i.e. $\,\gcd(x,a) = 1 = \gcd(y,b)$. $${\rm if}\ \ \dfrac{x}a+\dfrac{y}b=\dfrac{z}c\ \ {\rm does}\  \dfrac{ab}{\gcd(a,b)^2}\mid c\ ?\qquad$$
41202
Let $d=(a^n,b^n), d_1=(a,b)^n$; need show $d\mid d_1$ &amp; $d_1\mid d$ to prove.  For the general case of $(a,b)\ne 1, (a,b)= (d_1)^\frac{1}{n}$, else $(a,b)=d_1$.  Let, $d_2= (a,b)=d_1^\frac{1}{n}\implies a=a'd_2, b = b'd_2\implies a^n= (a')^nd_2^n, b^n= (b')^nd_2^n$  Unable to progress further.
102709
Prove that if $a, b$ are both even then $\gcd(a,b) = 2\cdot\gcd(a/2,b/2)$.   Little confused here. I have tried the following but it's basically just repeating the proof unfortunately:   $a = 2 \cdot a_1 $ and $ b = 2 \cdot b_1$ (Take factor of 2 out)  $d = \gcd(a,b)$  $d = 2\gcd(a_1, b_1)$ $a/2 = a_1$ and $b/2 = b_1$   I've even confused myself! Could someone help!
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.
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.
24761
Show that if $n$, $a$, $b$, and $c$ are positive integers with $\gcd(a, b) = 1$ and $ab = c^n$, then there are positive integers $d$, and $e$ such that $a = d^n$ and $b = e^n$.   I know that (by Bezout) $\gcd\left(a,b\right) = 1$ implies $ax + by = 1$ for some integers $x$ and $y$, and also that $\gcd\left(a^n,b^n\right) = 1$, but this does not help me.
8243
If $\gcd(a,b)=1$, how can I find the values that $\gcd(a+b,a^2+b^2)$ can possibly take? I can't find a way to use any of the elemental divisibility and gcd theorems to find them.
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 $a^2 ~~\&amp;~~ b^2$.
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.
100174
I'm just wondering the reason behind being able to do this:  $$\gcd(a,c) \cdot \gcd(b,c) = \gcd(ab,bc,ac,cc)$$  Is there an elementary proof to the reasoning behind this?