Prove the following proposition:    Suppose $a,b$ are fixed integers. Then $\{ ax+by\mid x,y\in\mathbb Z\} = \{ n(a,b) \mid n\in\mathbb Z\}$.
38221
I have the following problem:  Let $a, b \in\mathbb{Z}$. Show that $\,\{ ax + by\ :\ x, y \in \mathbb{Z}\} = \{ n \gcd(a,b)\ :\ n\in \mathbb{Z} \}$   I understand that the Bezout's lemma says that $gcd(a,b) = ax +by$, so Im not really how you would go about proving the above, it doesn't really make sense to me. Any help is appreciated1
16715
Let $A\in M_{n\times n}(\mathbb{R})$ symmetric and $b\in \mathbb{R}^n$.  How I can prove that if $b\perp \ker A$ then there exists $x\in \mathbb{R}^n$ such that $Ax=b$?  Many thanks!
19620
I was looking for an answer for this problem in terms of matrices, but I really don't know how to prove this result. The proposition says that:  Let $A\in M_{m\times k}(\mathbb{C})$, $B\in M_{k\times p}(\mathbb{C})$ and $C\in M_{p\times n}(\mathbb{C})$, then $\textrm{rank}(AB)+\textrm{rank}(BC)\leq \textrm{rank}(B)+\textrm{rank}(ABC)$
13650
I don't know how to explain or how to prove the following statement  If $n=ab$ and $a,b \in \mathbb{N}$ then $2^a-1 \mid 2^n-1$.  Any ideas? Perhaps an induction?  Thanks in advance.
515
The question is      Prove that $a\mathbb{Z}\cap b\mathbb{Z}=[a,b]\mathbb{Z}$.   Hint:First prove $b\mid a \Leftrightarrow a\mathbb Z\le b\mathbb Z$ and then prove $a\mathbb Z+b\mathbb Z=(a,b)\mathbb Z$   I managed to prove $b\mid a\Leftrightarrow a\mathbb Z\le b\mathbb Z$. About the next claim: From the first statement ($b\mid a...$) it follows that since $(a,b)\mid a$ and $(a,b)\mid b, a\mathbb Z+b\mathbb Z\le(a,b)\mathbb Z$. My problem is proving the opposite side (a.k.a $a\mathbb Z+b\mathbb Z\ge(a,b)\mathbb Z$). How can I do so and how can I proceed from $(a,b)$ to $[a,b]$?
57867
Let $x_1,...,x_n$ be distinct integers. Prove that  $$\prod_{i&lt;j} \frac{x_i-x_j}{i-j}\in \mathbb Z$$   I know there is a solution using determinant of a matrix, but I can't remember it now. Any help will be appreciated.
123678
Show that if $x, y, z$ are positive integers, then $(xy + 1)(yz + 1)(zx + 1)$ is a perfect square if and only if $xy + 1, yz + 1, zx+1$ are perfect squares.
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 $.
9789
Here is my proof of $\gcd(a,b)=ax+by$ for $a, b, x, y \in \mathbb{Z}$. Am I doing something wrong? Are there easier proofs?  $a,b \in \mathbb{Z}, g=\gcd(a,b)$ and suppose $g \neq ax + by$. Let $c$ be a common divisor of $a$ and $b$. Then   $$\forall x', y' \in \mathbb{Z}: c | ax' + by'\Longrightarrow\exists q_1, q_2 \in \mathbb{Z}\,\,\, s.t.\,\,\, c q_1 = ax' + by'\,\,,\,\,cq_2 = g$$   So   $$gcd(a, b)=g=c q_2 = c q_1 \frac{q_2}{q_1} = \left(\frac{q_2}{q_1}x'\right)a + \left(\frac{q_2}{q_1}y'\right)b$$ So if we have $q_1|q_2$ then we found $\gcd(a,b)=ax'+by'$ for all $x', y' \in \mathbb{Z}$.  Now   $$\frac{q_1}{q_2}=\frac{c q_1}{c q_2} = \frac{ax'+by'}{g}$$ but $g|a$ and $g|b$ so $\exists q_3 \in \mathbb{Z}: \frac{q_1}{q_2}=q_3 \Rightarrow q_1=q_2 q_3 \Rightarrow q_1 | q_2\,$ . QED.
30665
Suppose that $a,b,m,n$ are natural numbers, prove the following statement:  If $a({m \over n})=b$ and $(a,n)=1$, then ${m \over n} \in \mathbb{N}$
18783
Let $ a,b,c$ positive integer such that $ a + b + c \mid a^2 + b^2 + c^2$.   Show that $ a + b + c \mid a^n + b^n + c^n$ for infinitely many positive integer $ n$.   (problem composed by Laurentiu Panaitopol)  So far no idea.
69122
How can I quickly prove that if for every $x \in \mathbb{R}^N$ $$Ax=Bx$$ then $A=B$ ? Where $A,B\in \mathbb{R}^{N\times N}$. Normally I would multiply both sides by inverse of $x$, however vectors have no inverse, so I am not sure how to prove it.
58551
Form the assumption, we can say $b=ak$ ,$k$ integer, $a=bm$, $m$ integer.  Intuitively, this conjecture makes sense. But I can't make further step.
52713
The problem statement is: for $a,b$ coprime, prove that there exists positive integers $x,y$ such that $ax-by=1$.  The question is form Arthur Egnels' problem solving text, and from the chapter on the pigeonhole principle.    My question: His proof begins by generating a list $a,\ldots,a(b-1)$, where he then points out that each element in this list(which proceeds sequentially) doesn't have remainder $0\pmod{b}$.  He then shows that we arrive at a contradiction if we assume that we also don't get remainder $1$ in this list $\bmod b$.  To show the contradiction in this last statement, he first states that we would have positive integers $p,q $ where $ 0&lt;p&lt;q&lt;b$ so that $pa\equiv qa \pmod{b}$.  He then goes on to point out that since we $a$ and $b$ are coprime, we then have that $b| q-p$.  I understand the last part here, but don't see why we get $pa\equiv qa \pmod{b}$.  Thanks in advance
73192
Prove: If $a\mid m$ and $b\mid m$ and $\gcd(a,b)=1$ then $ab\mid m$  I thought that $m=ab$ but I was given a counterexample in a comment below.  So all I really know is $m=ax$ and $m=by$ for some $x,y \in \mathbb Z$. Also, $a$ and $b$ are relatively prime since $\gcd(a,b)=1$.   One of the comments suggests to use , i.e., $aq+br=1$ for some $q,r\in\mathbb{Z}$. Any more hints?  New to this divisibility/gcd stuff. Thanks in advance!
1089
Suppose $a$ and $b$ are positive integers, and that $d=\gcd(a,b)$. Suppose we have found integers $x$ and $y$ such that $ax+by=d$. Prove that $x$ and $y$ are relatively prime.
101000
Let $X\subset R$ and $Y \subset R$ where X and Y have finite cardinalities. Let also, $a,b \in R$. How to show that $|aX+bY|=|X||Y|$ almost everywhere (measure of $(a,b) \subset R^2$ such that $|aX+bY|&lt;|X||Y|$ is 0) ?  Where we define $|aX+bY|=\{ax+by:x \in X, y \in Y \}$.  More, specifically let $X=\{x=d_x*z: z \in \mathcal{Z}, d_x \in \mathcal{R}\}$ (so, it's a lattice. Where each point is $d_x$ away from one another.)  The above is easy if $X=Y$ then the proof of $|aX+bX|=|X||X|$ almost everywhere is fairly simple.   Now what if $Y$ is lattice with $d_y \neq d_x$. How to show it then? Thanks in advance.