Let $x,y$ be coprime integers. I want to prove that if $z \in \mathbb Z$ is a multiple of $x$ and $y$ then $z$ is a multiple of $xy$.   I know that $\exists r,s,m_1, m_2 \in \mathbb Z$ s.t. $$z=m_1x=m_2y$$ $$xr+ys = 1$$. But this is as far as I've gotten. I'm not sure how to proceed from here.
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?
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.
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!
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?
16747
I'm having problems with an exercise from Apostol's Introduction to Analytic Number Theory.     Given $x$ and $y$, let $m=ax+by$, $n=cx+dy$, where $ad-bc= \pm 1$. Prove that $(m,n)=(x,y)$.   I've tried to give a proof, but I suspect it's wrong (or at least not very good). I would be very thankful for any hints/help/advice!  My proof:     We observe that since $ad-bc= \pm 1$, $ad=bc \pm 1$, and $(ad,bc)=1$.   Now, $(a,b) \mid m$ and $(c,d) \mid n$ but      $$(a,b) = ((d,1)a,(c,1)b)=(ad,a,bc,b)=(ad,bc,a,b)=((ad,bc),a,b)=(1,a,b)=1.$$      Similarly, we determine $(c,d)=1$. So, $1=(a,b) \mid m$ and $1=(c,d) &gt; \mid n$. But $(x,y)$ also divide $m$ and $n$. Since $(x,y) \geq (a,b)=(c,d)=1$, this implies that $(x,y)=m,n$. Hence   $(m,n)=((x,y),(x,y))=(x,y)$.
101588
I am self studying number theory from Introduction to Analytic number theory by Apostol.   I have a doubt in one argument of proof of Theorem 5.14 and I am posting its image.      My doubt is in 2nd line of 2 nd paragraph how Apostol writes - If y is a solution of (5) then ay $\equiv $ at ( mod m) . How does y being a solution of (5) implies that?       Can someone please help
44650
I'm trying to prove that if there are to numbers $n,m$ (natural numbers), and their smallest common multiple is $k$, so that $k = n·i$ and $k = m·j$ for some $i,j$ natural numbers, any common multiple $q$ of $n,m$ is a multiple of $k$, so that there exists a $g$ so that $q=k·g$.  Logically it seems correct, but I can't seem to prove it... Any help?
35983
1) How do I prove the following:   Let A = {6a + 4b ∈ Z : a, b ∈ Z} and B = {2a ∈ Z : a ∈ Z}. Show that A = B.  Thank you all for the help!
103493
Suppose that $I$ and $J$ are ideals of $\mathbb{Z}$, with $I=(m)$ and $J=(n)$. This question has two parts:  1) Let r be the least common multiple of $m$ and $n$. Show that $I\cap J = (r)$.  2) Let $d=(m,n)$.  Show that $I+J = (d)$.
7242
I can't crack this one.      Prove: If $\gcd(a,b,c)=1$ then there exists $z$ such that $\gcd(az+b,c) = 1$ (the only constraint is that  $a,b,c,z \in \mathbb{Z}$ and $c\neq 0)$
38627
Let $a$ and $b$ be integers. Show that if $(a,b) = 1$ then $(a,a + b) = 1$.   I'm stuck on how to do this proof. I tried using the fact that $ax+by=1$ but that didn't seem to help.
106161
I have to prove that $\mathbb Z [X] / (X) \cong \mathbb Z[X] / (X+1) \cong \mathbb Z [X] / (X+2015)$.  I think that one answer could be that $\mathbb Z[X]/(X) \cong \mathbb Z(0)$, $\mathbb Z[X]/(X+1)\cong Z(-1)$ and $\mathbb Z [X] /(X+2015)\cong \mathbb Z(-2015)$, as $0, -1, -2015$ are roots of these polynomials. Also, as $0, -1, -2015\in \mathbb Z$, then all of them are isomorphic to $\mathbb Z $, hence isomorphic between them.  But I don't know very well how to justify this rigorously and my teacher has said to me that there is a simpler way to proof this, without using the roots of these polynomials.
41005
How would one prove that the ideal $(xy,xz,yz)$ of $k[x,y,z]$ for some field $k$, cannot be generated by two polynomials. In other words, prove:   $$(xy,xz,yz) \neq (f,g)\; \forall f,g \in k[x,y,z].$$   Furthermore, can somebody abstract this fact and prove it for $k[x_1,x_2,...,x_n]$? Maybe that the ideal of certain combinations of multiplied variables $x_i$ is not generated by less elements. Please ask if my question is unclear.  Edit: Specifically, I want to say that the ideal generated by all combinatorial products of $n-1$ variables in $k[x_1,x_2,...,x_n]$ cannot be generated by less elements. But if somebody has a different abstraction, please share.
51601
Let $x,y,z$ be 3 co-prime integers. Show that the only nontrivial solutions of: $$x+y+z=xyz$$ are only the permutations of $(x,y,z)=\pm(1,2,3)$
96058
I start with knowing that two numbers are coprime if:  $n*k + m*j = 1$  So, setting $k = a$ and j = $a+1$ I can solve as follows:  $n*a + m*a + a$  Then,  $a(n+m) + m = 1$  Where can I go from here?
38990
That is,     why is $\varphi(AB)=\varphi(A)\varphi(B)$, if $A$ and $B$ are two coprime positive integers?   It's not just a technical trouble—I can't see why this should be, intuitively: I bellyfeel that its multiplicativity should be an approximation at most. And why the minimum value for $\varphi (n)$ should be $ \frac{4n}{15} $ completely passes me by.
40803
For the sequence $a_n = 6^{2^n} + 1$:   How can I prove that the elements of this sequence are pairwise coprime, i.e. prove that if $m$ is not equal to $n$ then $\gcd(a_m, a_n) = 1$. (begin by proving that $a_n∣ (a_{n+1} - 2)$)