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I know that there are lots of questions with this format with duplicated answers:

  1. If $\gcd(a,b) = 1$, show that $\gcd(2a+b, a+2b)=1 \mbox{ or } 3$

  2. Show that $\gcd(a + b, a^2 + b^2) = 1$ or $2$ if $\gcd(a, b)=1$

  3. Prove $\gcd(a+b, a-b) = 1$ or $2\,$ if $\,\gcd(a,b) = 1$

There are a myriad of different solutions for these questions. I do not want to know exactly how to solve each of these problems. What I want to know is why the way I am solving these are apparently wrong. Every time I posted one of these problems with my resolution, people close my question and point me to duplicated questions and although this helps me solve a particular question, this does not help me learn, since people are not pointing out what I am doing wrong.

People close my questions as if I was lazy not to search for similar questions that have already been answered, but please realize it takes way more effort to show my work here. And if I am making this effort, is because the answers are not working for me. There is a difference between seeing the right way and understanding why your way is wrong. That being said what I am doing with this questions is:

Let

$$d_1 = \gcd(2a+b, a+2b)$$ $$d_2 = \gcd(a + b, a^2 + b^2)$$ $$d_3 = \gcd(a+b, a-b)$$

If $d_i$ is the $\gcd(x,y)$, than $d_i\mid x+y$ and $d_i\mid x\cdot y$, so:

$$d_1 \mid (2a+b) + (a+2b) = 3(a + b)$$ $$d_2 \mid a^2 + b^2 = (a+b)^2 - 2ab$$ $$d_3 \mid (a+b) + (a-b) = 2a$$ $$d_3 \mid (a+b) - (a-b) = 2b$$

As I understand, since $\gcd (a,b) = 1$, so $a$ and $b$ will not have prime factors in common which implies that $ab$, $a+b$ and $(a+b)^2$ will not have prime factors in common with $a$ or $b$. So I can just look for $d_i$ in the the factors that are not $ab$, $a+b$ and $(a+b)^2$. Which gives me:

$$d_1 \mid 3(a + b) \implies d_1 = 3$$ $$d_2 \mid 2ab \implies d_2 = 2$$ $$d_3 \mid 2a \implies d_3 = 2$$ $$d_3 \mid 2b \implies d_3 = 2$$

Is my argument sufficient to prove these problems? If not, why? I know the "proper way" to solve these questions, people have already pointed out all the solutions out there. I am interested to know, why my argument does not hold true.

Martin Sleziak
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user3347814
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    I feel that in all Stackexchange people dont value mistakes as a way to learn. They always point out the right way, they rarely point out why way you are doing things are wrong and that is just as valuable for learning. – user3347814 May 28 '22 at 16:42
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    In every case you are duplicating the error already explained in your prior question, i.e. generally to infer $,d\mid n c\Rightarrow d\mid n,$ you need to prove $(d,c) = 1$ and then apply Euclid's Lemma (and note the conclusion is $,d\mid n,,$ not $,d = n).\ $ Said question is the only one you have posted of this form, so your remark "every time I post ..." is puzzling. You do have other dupes, but not of this type. – Bill Dubuque May 28 '22 at 17:35
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    If you have questions after dupe closures, please (first) ask them in the question closed (or the linked dupes). and only post a new question if you get no satisfactory answers. – Bill Dubuque May 28 '22 at 17:38
  • All my questions regarding these kind of problems explain my thought process and ask for guidance, not exclusively an answer.. All of them, including this very question, were closed without an actual answer. Two lines of comments are not helping me. – user3347814 May 28 '22 at 17:39
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    Again, you have only one question of this sort. and the error has been explained. Please read the linked dupes to understand the many methods available. – Bill Dubuque May 28 '22 at 17:41
  • I do not know what to do. In this question, I explicitly cite the dupes and explain that they do not help me. – user3347814 May 28 '22 at 17:41
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    Why don't any of the linked standard answers help you? – Bill Dubuque May 28 '22 at 17:43
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    They all provide arguments that are vastly different than mine. The arguments are greatly better than mine, of course (and right, which is important), but they do not directly address why my arguments are wrong and what should I do to improve them. I could easily just memorize their solution, but that is not learning. And the fact that my questions are getting closed before they can be answered is not helping. You clearly care for this community and really know your math. Why not just answer the question, instead of arguing about closing it? – user3347814 May 28 '22 at 17:50
  • I do not understand why for d1∣3(a+b)⟹d1=3 the fact that gcd(a,b) = 1 is not sufficient. – user3347814 May 28 '22 at 17:51
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    What is your proof that it is sufficient? The linked arguments are not "vastly different" They already show correctly how to do what you attempt to do. – Bill Dubuque May 28 '22 at 17:56
  • Let us continue this discussion in chat. – user3347814 May 28 '22 at 17:58
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    Please don't use chat - it has many defiecits, e.g. poor math display, not searchable / indexed, etc. So many users do not read / reply to chats. – Bill Dubuque May 28 '22 at 17:59
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    I feel that this is the biggest problem with stackexchange in general. People who know things seem to think that everything they know it obvious. But if I took my time to post a question, it is not obvious to me. Besides, I am working with the constrictions the amount of math that I have learnt so far. Closing questions just discourages me from posting questions. – user3347814 May 28 '22 at 18:07
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    Please continue in your prior question and explain precisely what remains "not obvious" to you, after my comment there. – Bill Dubuque May 28 '22 at 18:10
  • As for my doubts, take d1 as an exemple: this is true: d1∣3(a+b) and d1= 3 is also true, d1| 3 and d1 does not divide (a+b) is also true. I can not see, what I am doing wrong. Since every statement is true. – user3347814 May 28 '22 at 18:11
  • I am not arguing that I am right. I know I am not. I know my math skills are garbage, but I do not know enough math to see where my mistakes are. I am not trying to defend my answer, I am just trying to learn how to do better. – user3347814 May 28 '22 at 18:16
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    No, $,d_1 = 3,$ is not true. Rather, you claim it is but have not given a complete / correct proof. It seems you may also have misunderstandings about logic (how to prove implications), besides number theory. – Bill Dubuque May 28 '22 at 18:22
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    Yes, I do have misunderstandings about many thing in math. That is the reason I post questions, in hopes that someone will help me clarify this misunderstandings. – user3347814 May 28 '22 at 18:25
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    Closing my questions makes it hard for people to help me with this misunderstandings. – user3347814 May 28 '22 at 18:26
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    Talking purely about your approach in this comment : you're nearly correct on the first question. For example, when you reach $d_1 | 3(a+b)$ (which is correct),then you reject $a+b$ on the basis of coprimality (which is correct again), but that doesn't give you $d_1$ equal to $3$ : rather, it only gives you that $d_1$ divides $3$. $d_1=3$ isn't true all the time, for example when $a=1,b=2$, say. The other deductions are wrong : of course $ab$ shares factors with $a$ and $b$, if you think about it, so you don't have coprimality. You will have to be cleverer in those questions. – Sarvesh Ravichandran Iyer May 29 '22 at 11:23

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