First of all, I want to acknowledge that I had many of my questions closed because of duplicates. I am aware that this problem has been posted many times(here, here,here). I am not trying to get a solution to this problem, I want to understand which concepts I am not fully grasping.
What I am trying to do with this problem is applying the same logic that I saw on the answer for this question. For convenience I will post the question and its solution here:
Prove $\gcd(a+b, a-b) = 1$ or $2\,$ if $\,\gcd(a,b) = 1$
The accepted answer is:
Let $d$ be a common divisor of $a+b$ and $a-b$, then $d$ divides their sum $2a$ and difference $2b$. If a number divides two numbers it also divides their gcd, thus $d$ divides $2\gcd(a,b) = 2$. That implies that every divisor (including the greatest common divisor) is a divisor of $2$.
The same argument again in symbols:
Let $d \mid a+b, a-b$, then $d \mid (a+b)+(a-b) = 2a$ and $d \mid > (a+b)-(a-b) = 2b$ so $d \mid \gcd(2a,2b) = 2$.
So I want to apply the same logic to solve this question:
prove if $\gcd(a,b) = 1$, then $\gcd(2a+b,a+2b) \in \{1,3\}$
$$\text{Let }\gcd(2a+b,a+2b) = d \implies d \mid 2a+b\text{ and } d \mid a+2b$$
Using the same argument from the other question i.e if $z \mid x$ and $z \mid y$ $\implies$ $z\mid (x+y)$ and $z\mid (x-y)$:
$$ d \mid (2a+b) + (a+2b) = 3a+3b$$ $$ d \mid (2a+b) - (a+2b) = a-b$$
Using the same argument again:
$$ d \mid (3a+3b) + 3(a-b) = 6a$$ $$ d \mid (3a+3b) - 3(a-b) = 6b$$
Then using the final argument
If a number divides two numbers it also divides their gcd
$$d\mid gdc(6a,6b) = 6gcd(a,b) = 6$$
Finally if $d \mid 6 \implies d \in \{1,2,3\}$
I know this is not the right answer. I can not find my mistake. I am clearly missing some key concepts. For the life of me, I can not figure out what those are. Can someone please explain to me, why this approach works for one problem, but not for the other?
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 statement. However, a stronger true statement is that $\gcd(2a+b,a+2b)\in\{1,3\}$. You can obtain this by looking, not at $(2a+b)+(a+2b)$ and $(2a+b)-(a+2b)$, but instead at $2(2a+b)-(a+2b)=3a$ and $-(2a+b)+2(a+2b)=3b$.
There is nothing special about $x+y$ and $x-y$: if $d$ divides both $x$ and $y$ then $d$ divides every integer linear combination $ax+by$ of them ($a,b\in\Bbb Z$). Depending on what $x$ and $y$ are, some linear combinations might give stronger information than others.
(It's also worth noting that examples like $a=1,b=0$ and $a=b=1$ show that both $\gcd(2a+b,a+2b)=1$ and $\gcd(2a+b,a+2b)=3$ are possible; so that stronger true statement is as strong as one could hope for.)