Let $a, m, n\in\mathbb{N^{*}}$  With $m&gt;n$. Show $${{(a^{2^m}-1,a^{2^n}+1)=a^{2^n}+1}}$$$$$$My thoughts: If we could show that $$(a^{2^n}+1)\mid (a^{2^m}-1)$$ the property that gcd says "If  $\;a\mid b\Longrightarrow (a,b)=a$" I can not work well with two unknowns in number theory.
67668
I've been struggling with this all day today. I imagine it's not very hard, but my algebra skills are terrible. So, how can I show that if $m&gt;n$ and $a$ is a positive integer, then $$a^{2^n}+1 \mid a^{2^m}-1.$$ I just can't get a coherent picture of what I'm supposed to do. If this too (see yesterday's question) boils down to the Unique Factorization Theorem, I would be very grateful for any intuition on how to think about this!
38165
Suppose $\forall (a;b;c;d) \in \mathbb{Z}^4$ as $\gcd(a,b)=\gcd(c,d)=1$:  How can I prove that $\gcd(a,bc)=\gcd(a,c)$?  Also, how to prove that $\gcd(ac,bd) = (\gcd(a,d))*(\gcd(b,c))$?
18138
I would like to prove that  $$a^{2^n}\equiv 1 \pmod {2^{n+2}}$$  I tried induction but could not get it.  Thank you very much!
72907
Prove that for any even positive integer $n$, $n^2-1 \mid 2^{n!}-1$     This is from a book. They have given the proof. But I didn't understand it well. I am looking for a simpler proof. Or it will be helpful if someone explain this a little bit more - Proof: Let $m = n+1$ then we need to prove that $m(m-2) \mid 2^{(m-1)!}-1$. Because of $\phi(m) \mid (m-1)!$, we have $2^{\phi(m)} -1 \mid 2^{(m-1)!} -1$. And from Euler's theorem, $(m-2) \mid 2^{(m-1)!}-1$. Because $m$ is odd, $gcd(m,m-2)=1$ and the conclusion follows.
74959
Prove or disprove 'If $\gcd(a,b)=1$ then, $\gcd(a^2,b^2)=1$, with $a,b\not= 0$'  I need to prove this statement. I think it is true and also the converse is true.  I took some examples such as $\gcd(2,5)=1$, and $\gcd(4,25)=1$...  But how can I prove this? I know that $1=ax+by$ for some $4x,y\in \mathbb{Z}$.. But not sure   How to go from there..   Any clue?
114600
I'm looking for solution for this problem     Let $a$ be an integer greater then 1. Then, the set   $A\mathrel{:=}\{a^n(a+1)-1\mid n\in\mathbb{Z}_{+}\}$ a has an infinite   subset b such that it's elements are pairwise coprime.   I found similar question  but proposed solutions are wrong:  Author of solution claims that " Then $d$ is also a multiple of $m$ and using the well-known fact that $x−y$ divides $x^k−y^k$, we see that $a^m−1$ is a multiple of $a^d−1$ "  Here we can find a small but critical mistake: from $m\mid d$ follows that $a^m-1 \mid a^d-1$ but no $a^d-1\mid a^m-1$.  Also author states that if $m\mid n$ that $\gcd(X_m,X_n)=1$. It's also false because we can take $X_1=a(a+1)-1$ and it's prime factor $p$. But then $X_p=a^p(a+1)-1$ is also divided by $p$ because of Fermat's little theorem.
110473
Let $g_n= {2^2}^n +1 $. Prove $g_0 · g_1 · · · g_{n−1} = g_{n} − 2$.    I'm not sure how to start this proof, if it should be done algebraically or if I should try to use a proof by induction.
33423
I want to show that for $\gcd(a,b) = 1$ $a,b \in Z$   $\gcd(a+b, a-b) = 1$ or $\gcd(a+b, a-b) = 2$ holds.      I think the first step should look something like this:      $d = \gcd(a+b, a-b) = \gcd(2a, a-b)$   From here I tried to proceed with two cases. 1: $a-b$ is even, which leads to $\gcd(a+b, a-b) = 2$ 2: $a-b$ is odd, which leads to $\gcd(a+b, a-b) = 1$  My main problem I think is, that I do not know how I should include $\gcd(a,b) = 1$ in the proof.  Any help is appreciated. Thx in advance. Cherio Woltan
25374
I'm trying to prove by induction that for all $a,b \in \mathbb{Z}$ and $n \in \mathbb{N}$, that $(a-b) \mid (a^n-b^n)$. The base case was trivial, so I started by assuming that $(a-b) \mid (a^n-b^n)$. But I found that: \begin{align*} (a-b)(a^{n-1}+a^{n-2}b + a^{n-3}b^2+...+b^{n-1}) &amp;= a^n-b^n. \end{align*}  Doesn't this imply that $(a-b) \mid (a^n-b^n)$ as $a^{n-1}+a^{n-2}b+\dots+b^{n-1}$ is clearly an integer? This obviously isn't a proof by induction, but is there anything wrong with taking this approach to prove this result, other than the fact that it isn't what is being asked?   Aside, I'm having trouble getting started with the proof by induction. I've tried making the above equivalence useful during the induction step, but it doesn't seem to be helpful, so any hints would be greatly appreciated!
55143
The problem's quite clear. Prove that $$\gcd(a^2, b^2) = \gcd(a, b)^2$$  This is easy to understand intuitively and using the Fundamental Theorem of Arithmetic would be easy but I want to prove it by using the Divisibility axioms and the GCD. I have attempted it and am mentioning my results.  Here $d = \gcd(a, b)$.   $$d | ax + by \implies d^2 | a^2x^2 + 2axby + b^2y^2 \implies d^2 | a^2x^2 + b^2y^2 \implies \gcd(a^2, b^2) | d^2$$  I feel that this is not necessary. Please help.  Also, the proof can be generalized to $$\gcd(a^n, b^n) = \gcd(a, b)^n$$  Any help would be appreciated.
21280
I just want to know if my proof to this question is correct. First, I proved it was true for $n = 4$. $$4!&gt;2^4$$ $$24&gt;16$$ Then, I assumed that it was true for $n=k$. $$k!&gt;2^k$$ Afterwards, I tried to prove it for $n=k+1$. $$(k+1)!&gt;2^{k+1}$$ $$(k+1)k!&gt;(2)(2^k)$$ Since $k!&gt;2^k$, am I allowed to simply do $$(k+1)(2^k)&gt;(2)(2^k)$$ $$(k+1)&gt;2$$ and this is true for all $k\geq3$. Would the last two steps be illegal because I only assumed that $k!&gt;2^k$ is true?
23279
The problem is to prove the following   If $\gcd(a,b) = c$, then $\gcd(a^m, b^m) = c^m$    I know that this can be solved easily by proving that $c\mid a \implies c^m \mid a^m$ and $c\mid b \implies c^m \mid b^m$. So the greatest common divisor of $a^m$ and $b^m$ is $c^m$. But the problem is that I want to use Bézout's Lemma (Identity) to prove this.  I first introduce the following notation:  $$a = ca' \quad \quad b = cb', \quad \quad \text{where $(a',b') = 1$.}$$  Now from Bézout's Lemma, becaus $a'$ and $b'$ are coprime integers then we have integer solutions for the following equation:  $$a'x + b'y = 1$$  It's obvious that also the following equation has an integer solution:  $$a'^mx_1 + b'^my_1 = 1$$  Although I'm not able to prove this using Bézout's Lemma.  Now multiply both sides by $c^m$ we get:  $$(a'c)^mx + (b'c)^my = c^m$$ $$a^mx + b^my = c^m$$  But Bézout's Lemma states that for given $c^m$ is the greatest common multiplier of $a^m$ and $b^m$ or it is a multiplier of it.  But how to prove that it's not a multiplier of the greatest common multiper, and in fact is the greatest common multiplier. Is it possible to solve this problem with Bézout's Lemma?
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.
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 $.
41621
Is it possible for some integer $n&gt;1$ that $2^n-1\mid 3^n-1$ ?   I have tried many things, but nothing worked.
24893
Prove by induction that $\forall n \in\mathbb N$, $3\mid4^{n}-1$.  1) For $n = 1$ the statement is obviously true.  2) Now what about $n+1$? I was thinking of writing $4^{n}-1$ as $2^{2n}-1$ and then $4^{n+1}-1 = 2^{2n+2}-1$ but that got me nowhere.
113764
I'm stuck at the following question: let $d$ divide $2^{2^n} +1 $. Show that $d \equiv 1 \pmod{2^{n+1}}$.  All I've managed to do so far was to notice that $2^{n+1} \mid 2^{2^n}$ for every natural $n$, so if $dl = 2^{2^n} + 1$, we have $dl \equiv 1 \pmod{2^{n+1}}$ - which merely states that $d$ is invertible, which is trivial since it is odd. This is homework, are there any hints?