4
votes
What I am doing wrong when trying to solve: if $\gcd(a,b) = 1$, then $\gcd(2a+b,a+2b) \in \{1,3\}$?
First, $d\mid 6$ (and $d>0$) implies $d\in\{1,2,3,6\}$ rather than $d\in\{1,2,3\}$.
Your argument is a correct proof that if $\gcd(a,b)=1$ then $\gcd(2a+b,a+2b)\in\{1,2,3,6\}$, which is a true ...
- 64.3k
2
votes
What I am doing wrong when trying to solve: if $\gcd(a,b) = 1$, then $\gcd(2a+b,a+2b) \in \{1,3\}$?
You haven't made any logical mistakes, and you have in fact proved a true statement: it is certainly the case that if $gcd(a,b)=1$ then $gcd(2a+b,2b+a) \in \{1,2,3\}$. The only reason you aren't ...
- 4,441
1
vote
Number Theory : $\frac{x + y }{\gcd(x,y)} \geq q$
This problem is harder and more interesting than it has gotten credit for. EDIT: We first establish "only if" direction. And then below, we establish the "if" direction:
The "...
- 16.1k
1
vote
Prove that either $m$ divides $n$ or $n$ divides $m$ given that $\operatorname{lcm}(m,n) + \operatorname{gcd}(m,n) = m + n$?
If $m=n$ then the equation is true and the statement to be proved is true.
If $m\ne n$, label the numbers so that $m>n$. Then, modulo $m$, $\gcd(m, n)=\textrm{LHS}=\textrm{RHS}=n$. But $0<\gcd(m,...
- 2,463
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