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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?

Arturo Magidin
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user3347814
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    When you write something like "if $d\mid 6\implies d\in{1,2,3}$", it reads as if the premise is the implication, and you have failed to provide the consequent. Either replace "$\implies$" with "then", or remove the "if". What you seem to want is for us to read is $(\text{if }d\mid 6)\implies (d\in{1,2,3})$, which does not parse. Formulas should make sense when you read them out loud, and $p\implies q$ is either pronounced "if $p$ then $q$", or "$p$ implies $q$". – Arturo Magidin May 28 '22 at 21:14
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    Small logical mistake in going from $d\mid 6$ to $d\in{1,2,3}$ (missing $6$), but that is neither here nor there... – Arturo Magidin May 28 '22 at 21:21

2 Answers2

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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.)

Greg Martin
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  • You are saying that there is a stronger statement than mine, but how do I know which one is stronger? If $d \mid 3$ is stronger than $d \mid 6$, what garanties that there are no arguments stronger than $d \mid 3$? – user3347814 May 28 '22 at 21:30
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    See my last paragraph. – Greg Martin May 28 '22 at 21:45
  • By that logic, this answer https://math.stackexchange.com/questions/32737/prove-gcdab-a-b-1-or-2-if-gcda-b-1/32739#32739 should have provided the same argument and verify that {1,2} were valid answers, right? – user3347814 May 28 '22 at 21:47
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    That answer successfully argues that its gcd must divide 2. If one wanted the additional information that all divisors of 2 were possible, then that would require an additional verification. I suggest being very conscious and explicit (for a while until it becomes second nature) about what statement is being proved: an if-then statement is a perfectly valid mathematical fact regardless of whether some instances of the "then" part never happen in the presence of the "if" part (in other words, regardless of whether the converse of the if-then statement is true). – Greg Martin May 29 '22 at 01:18
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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 satisfied with it is that you know the question wants you to prove an even stronger statement, that $gcd(2a+b,2b+a) \in \{1,3\}$.

All we have to do to get from what you've proved to what you want to prove is to show that the gcd can't be $2$. We can do that as follows: suppose for contradiction that $gcd(2a+b,2b+a)=2$. Then $2|2a+b$ and $2|2b+a$...

Can you see how to get from that to a contradiction?

Chessanator
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  • Given that one of my mistakes was that $d \in {1,2,3,6}$, I would have to do the same thing for 6 as well, right? Could you show me how, I do not see the contradiction for both 2 and 6. Furthermore. How do I show that there are no contradictions with 1 and 3? – user3347814 May 28 '22 at 21:44