prove or disprove :  Suppose that $a$, $m$ and $n$ are positive integers such that $(m, n)=1$. If $m\mid a$ and $n\mid a$, then $mn\mid a$.  I think it is true statement.  Let $m\mid a$ and $n\mid a$ then $mn\mid na$ and $nm\mid ma$ since $(m,n)=1$ then $mn\mid a$.  Ut is short answer and I am not sure if that is it complete answer?
3974
Using divisibility theorems, prove that if $\gcd(a,b)=1$ and $a|c$ and $b|c$, then $ab|c$.   This is pretty clear by UPF, but I'm having some trouble proving it using divisibility theorems. I was looking at , and if this could be proven, then the solution would follow immediately.
76898
Let $m$ and $n$ be two nonnegative integers. Assume that there is a group isomorphism $\mathbb{Z}^m \cong \mathbb{Z}^n$. Prove that $m = n$.   I tried using a contrapositive, ($m \neq n$ implies $\mathbb{Z}^m \ncong \mathbb{Z}^n$), and I think the problem is that there won't be a homomorphism, but did not get anywhere. Is there a better approach to this problem?
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   $(m,n)=((x,y),(x,y))=(x,y)$.
109068
Find all positive integers $n$ such that $n$ is divisible by all the positive integers less than or equal to $\sqrt{n}$  My thought: If n is a positive integer, let d(n) denote the number of positive divisors of n and I will find n such that $d(n) \ge 2\sqrt{n} - 1$
120815
So, here is the question:  For any position integer $n$, let $T(n)$ be the number 1 if $n&lt;4$ and the number $T(n-1) + T(n-2) + T(n-3)$ if $n \geq 4$.  We have $T(1)=1, T(2)=2, T(3)=3$ $$T(4)=T(3)+T(2)+T(1) = 1+1+1+1 = 3$$  $$T(5) = T(4)+T(3)+T(2) = 3+1+1 = 5$$  Prove that: (Universal n is an element of all positive integers), $T(n)&lt;2^n$  Any suggestions?
27132
I want to prove that if $d \mid n \in \mathbb{N}$, then $\varphi(d) \mid \varphi(n)$.  It's given that $d \mid n$, so we know that $n = dm$, for some $m \in \mathbb{Z}$. Now, I want to show that $\varphi(d) \mid \varphi(n)$, i.e. that $\varphi(n) = k\varphi(d)$, for some $k \in \mathbb{Z}$. Drawing a blank of where to go from here.
123714
Question: Let $m,n\in\mathbb{N}$, prove that if $n|m$, $F_n|F_m$.  I've tried to use induction, but I don't really know where to start since there's $2$ numbers:  $n$ and $m\ \dots$ I did induction before with just $1$ number, like proving $1+2+\cdots+n=\frac{n(n+1)}{2}$, that's only dealing with one number $n$, but with this I have no clue how to do it.  Can anyone give me a little hint on how to start, I don't want you to do whole problem for me but can I have some hint so I know where to start?  Thanks for the help!
88366
Here $n$ belongs to natural numbers. Firstly, I proved the relation by putting $n = 1$ . Then, taking $$|\sin(mx)| \leq m|\sin x|$$ true, I had to prove $$|\sin(m + 1)x| \leq (m + 1)|\sin x|$$  Now, here I got stuck. How to prove it?? Please help.
88412
Given $51$ integers $a_1, a_2, \ldots, a_{51}$ satisfying $1 \leq a_i \leq 100$. Prove that there exist $i$ and $j$ such that $a_i \mid a_j$.   I'm a little unsure how to approach this problem. Thanks for any help.
61352
If $n$ be a positive integer greater than $1$, then prove that $$2^n&gt;1+n\sqrt{2^{n-1}}$$    I found this problem under $AM \ge GM$ chapter. Help me to solve this problem using  $AM \ge GM$. Thanks in advance.
88669
Let $n$ and $r$ be positive integers with $n \ge r$. Prove that   Still a beginner here. Need to learn formatting.  I am guessing by induction? Not sure what or how to go forward with this.  Need help with the proof.
17137
In respect to a larger proof I need to prove that $(n-m) \mid (n^r - m^r) $ (where $\mid$ means divides, i.e., $a \mid b$ means that $b$ modulus $a$ = $0$). I have played around with this for a while now but to no avail and would be grateful if you could help me.
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.
67794
I am not sure if I am doing this right, but I have this:  There exists an integer $k$. $2k =$ positive number  $(2k)^3 - 2k$ [*And this is where I get lost. How does one prove this?]
13989
I'm not sure I've got this right. When proving $a^n \mid b^n \Rightarrow a \mid b$, can we do this indirectly? In short,  "Suppose $a$ does not divide $b$, this implies that $a^n$ does not divide $b^n$. But $a^n \mid b^n$, hence $a$ divides $b$."  How about $n^n \mid m^m \Rightarrow n \mid m$? Can we do this the same way?
121073
If $m,n\in N$ Prove that there is such a positive integer k, such that $(\sqrt{m}+\sqrt{m+1})^n=\sqrt{k}+\sqrt{k+1}$ I attempted to solve this question using binomial coefficients, saying: $(\sqrt{m}+\sqrt{m+1})^n=\sqrt{k}+\sqrt{k+1}=\sum\limits_{a=0}^n {n\choose a}*\sqrt{m}^{n-a}*\sqrt{m+1}^a$ and from here I was thinking that I had to do something with $\sqrt{m}^{n-a}$ with $\sqrt{k}$ and do something else with $\sqrt{k+1}$ and $\sqrt{m+1}^a$. Unfortunately I couldn't think of what to do with this circumstance. I then thought of maybe taking $(\sqrt{k}+\sqrt{k+1})^2=(\sum\limits_{a=0}^n {n\choose a}*\sqrt{m}^{n-a}*\sqrt{m+1}^a)^2$, but this immediately over-complicated the question. Could you please explain to me how to solve this question and how to solve similar questions in the future?
22704
Prove that $\gcd(2^{2^m}+1,2^{2^n}+1)=1$ if $m,n$ are positive integers.  Let $d=\gcd(2^{2^m}+1,2^{2^n}+1)$, then $d\mid 2^{2^m}+1$ and $d\mid2^{2^n}+1$ and then $d\mid2^{2^m}+1-2^{2^n}-1$, i.e. $d\mid2^{2^m}-2^{2^n}$ where we have taken $m&gt;n$. Thus $d\mid2^{2^n}(2^{2^{m-n}}-1)$, but $d \nmid 2^{2^n}$ thus $d\mid2^{2^{m-n}}-1$.  Then how will I proceed??