Let $a,b,c$ be integers such that $\gcd(a,6)=3=\gcd(b,6)$. Show that $\gcd(a+b,6)=6$. Attempt: Let $d = \gcd(a+b,6)$. Then, by definition, $d \mid (a+b)$ and $d \mid 6$. It means that $d=1,2,3$ or $6$. On the other hand, we have $3\mid a$ and $3 \mid b$. Write $a = 3m$ and $b=3n$ for some integers $m$ and $n$. Then, $a+b=3m+3n=3(m+n)$ and so $3 \mid (a+b)$. But, $d \mid (a+b)$. Now, $a+b = dp$ for some integer $p$. Thus, we have $3 \mid dp$ which means $3 \mid d$ or $3 \mid p$. If $3 \mid d$, then $d=3k$ for some integer $k$. Hence, $d$ is either $3$ or $6$. But, how to get $d=6$?
123191
I was working my way through some number theoretic proofs and being a newbie am stuck on this problem :     If $(x, 4) = 2$ and $(y, 4) =2$, then $(x + y, 4) = 4$,     where $(a,b)$ denotes the greatest common divisor of $a$ and $b$.   My Solution  (Incorrect)   $x-4 = 2t$ $y-4 = 2p$ $x + y - 8 = 2(t+p) \Rightarrow x + y - 4 = 2f + 4 = 2(f+2)\Rightarrow 2$ divides $(x+y , 4)$   My Question:  My solution is definitely inadequate. Can someone help me out?
2131
I know that $\gcd(a,b)$ divides $a$ and $b$, and must also then divide $(a)(t)$ ($t$ being some integer).  This makes sense to me, but how do I prove it?  It seems that the addition of $(a)(t)$ is a continuation of the linear combination of $\gcd(a,b) = av + bu$ for some $v$, $u$ being elements of $\mathbb{Z}$.    Any help?
109119
Prove that if $a$ and $b$ are odd, coprime numbers, then $\gcd(2^a +1, 2^b +1) = 3$.   I was thinking among the lines of:  Since $a$ and $b$ are coprime numbers, $\gcd(a,b)=1$. Then there exist integers $x$ and $y$ such that, $ax+by=1$.   Then, $a=(1-by)/x$, $b=(1-ax)/y$  So if I write $2^a+1$ as:  $2^{(1-by)/x}+1$  Then can I say that the above expression is equivalent to 0 (mod 3)?
65951
How can I find $\gcd(a^m+1,a^n+1)$ with $a,m,n$ positive integers?    I have this idea:  Let $d=\gcd(m,n)$. Then there exist positive integers $x,y$ such that $mx-ny=d$ (WLOG). We shall find $G=\gcd(a^m+1,a^n+1)$.  If $m,n$ are odd, then $d$ is odd, therefore one and only one of $x,y$ is even. We have:  $$a^{ny}(a^d+1)=a^{mx}+a^{ny}=(a^{mx}-1)+(a^{ny}+1)=(a^{mx}+1)+(a^{ny}-1).$$  If $x$ is even and $y$ is odd, then $a^{m}+1\mid a^{mx}-1$ and $a^{n}+1\mid a^{ny}+1$, therefore $G\mid a^{ny}(a^d+1)$. If $x$ is odd and $y$ is even, then $a^{m}+1\mid a^{mx}+1$ and $a^{n}+1\mid a^{ny}-1$, thus $G\mid a^{ny}(a^d+1)$. However, since $\gcd(a^m+1,a^{ny})=\gcd(a^n+1,a^{ny})=1$, so $\gcd(G,a^{ny})=1$, hence $G\mid a^d+1$. We also have $a^d+1\mid a^{m}+1$ and $a^d+1\mid a^{n}+1$, so $a^d+1\mid G$. Thus $G=a^d+1$.  If $v_2(m)=v_2(n)=v_2(d)=k&gt;1$, then there exist some odd numbers $m_1,n_1,d_1$ such that $m=2^km_1,n=2^kn_1,d=2^kd_1$. We shall have $m_1x-n_1y=d_1$, so one and only one of $x,y$ is even, and we can use the same argument when $m,n$ are odd, so $G=a^d+1$.  However, if $v_2(m) \neq v_2(n)$, I cannot find any solutions for this. I think that $G \in \{1,2\}$, but I cannot prove or disprove it. How can I find $G=\gcd(a^m+1,a^n+1)$ if $v_2(m) \neq v_2(n)$ ? Moreover, is there anything from my arguments that can be improved ?
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??
7067
Show that if $a$, $b$ are positive integers, then we have: $\gcd(a,b) = \gcd(a + b, \mathrm{lcm}[a,b])$.
41678
As stated in the title, the problem to prove is  Let $a,b,c \in \mathbb{Z}$. If $\gcd(a,c)=1=\gcd(b,c)$, then $\gcd(ab,c)=1$.  I think I've proved it, but I would like a second opinion. Here goes:  PROOF Suppose $\exists d \neq 1 \in \mathbb{Z}$ s.t $\gcd(ab,c)=d$. It follows then that $d|ab$ and $d|c$. Now, if $d|a$ but not $b$, then since $d|c$, by linearity we have $d|\gcd(a,c)$. But $\gcd(a,c)=1$ so then $d=1$. The argument for $d|b$ is similar, resulting in $d=1$. Thus, since both cases lead us to $d=1$, we have a contradiction on our hands $\implies$ $ab$ and $c$ must be relatively prime $\implies \gcd(ab,c)=1$. $Q.E.D$  The question: was this a valid argument or do you think I should provide more?
18363
I need to prove that $z\gcd(a,b)=\gcd(za,zb)$.    I tried a lot, for example, looking at set of common divisors of the two sides, but I can't conclude anything from that. Can you please give me some advice how I can handle this problem? And $a,b,z \in \mathbb{Z}$.
7614
How can I prove that if $\gcd(a, b) = 1$, then $\gcd(ab, c) = \gcd(a, c) \times \gcd(b, c)$?  By eea there exists $ax+by=1$ from $\gcd(a,b)=1$ so a and be are co-primes there also exists $dk=a$ and $dj= b$ where $d=\gcd(a,b)=1$ this is all the information I have gathered from the question but I dont know how to approach and solve it. Can anyone help explain to me how to arrive at the answer? Thanks!
9261
Hi I'm practicing some number theory and I've come across the following problem,  Suppose $n \geq 1$ is an odd integer that divides $a^2 - 1$ for some integer $a &gt; 1$. Prove that,  $$n = \gcd(a-1, n)\gcd(a+1, n)$$  Now I know that the $\gcd(a-1, a+1) = 1$ if $a$ is even and $2$ if $a$ is odd and I believe this is relevant but I can't solve the problem. I've not really any idea where to start so even just help with that would be great.  Thanks
45763
Let $p$ be a prime. Suppose that $\gcd(a, b) = p$. Find $\gcd(a^2,b)$  for all integers a and b.   I was able to prove that $\gcd(a^2,b^2) = p$ but I don't know if it helps.  Any help is greatly appreciated.
89643
Prove that $\gcd(3^n-2,2^n-3)=1$ if and only if $\gcd(5,2^n-3)=1$ where $n$ is a natural number.   I didn't see an easy way to prove this using the Euclidean algorithm, but it seems true that both gcd's are not $1$ only if $n = 3+4k$. Is there an easy way to prove the statement?
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.
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!
110279
If integers $a$ and $b$ are not both zero, then $\gcd(a,b)=\gcd(a-b,b)$   Assume either $a=0$ or $b=0$. If $a=0$, then $\gcd(a,b)=\gcd(0,b)=b$ and $\gcd(a-b,b)=\gcd(-b,b)=-b$, so $\gcd(a,b)\neq\gcd(a-b,b)$.  If $b=0$, then $\gcd(a,b)=\gcd(a,0)=0$ and $\gcd(-b,b)=\gcd(a,0)=0$.  So, what can I say? Can you help? Can you check my proof-trying?
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.
2945
I tried to start by showing that $ \frac{a}{\gcd(a,b)} $ is always an integer, let's call it $d$, because $a$ is always a multiple of $(a,b)$ based on the definition of a g.c.d. I then tried to show that if $a | b$ then $a|bc$ from the theorem that states that $a|b$ implies that $a|bc$ for any integer $c$, but I cant seem to prove that $a | b$ or that $a|bc$ only if $d \mid c$ (where $d = \frac{a}{\gcd(a,b)}$ ). Any help would be much appreciated.