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Need the use of property of addition and subtraction of $\gcd$ for my approach to prove :
$(a,b)=1\implies (ab, a+b)=1$

My approach:
$(a,b)=1\implies (ab,b^2)=b$
$(a,b)=1\implies (ab,a^2)=a$

$((ab,b^2)=b\wedge (ab,a^2)=a)\implies (2ab,a^2+b^2)=b+a\implies (2ab,a^2+b^2+2ab) = a+b \implies (2ab,(a+b)^2) = a+b$

Need help to proceed further, if correct.

Of course it's not me
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jiten
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    The implication $((ab,b^2)=b∧(ab,a^2)=a)⟹(2ab,a^2+b^2) = b+a$ doesn't hold in general... I think it almost never holds actually. – mike239x Mar 07 '18 at 13:54
  • @mike239x Please provide details for the cause of invalidity of the addition operation on $\gcd$. – jiten Mar 07 '18 at 14:53
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    OK, underlying you have that from $x | y$ and $s | t$ you can't conclude that $x+s\ |\ y + t$. And now you'll ask the cause for that, I assume. – mike239x Mar 07 '18 at 15:01
  • @mike239x Please tell the reason. – jiten Mar 07 '18 at 15:06
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    I honestly don't know what to say. It just breaks way too easy. For example you take $y = 0$ and (since all numbers divide zero) you get that all numbers divide $t$... – mike239x Mar 07 '18 at 15:14
  • @mike239x Thanks, but then it points to invalidity of division of linear combinations too. Anyway, the explanation is too simple and is very much valid. – jiten Mar 07 '18 at 15:27

1 Answers1

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Hint :

  • if $(a,b)=1$, then $(c, ab) = (c,a)\times (c,b)$
  • $(a,b) = (a+b, b)$

Note that you can't "add" gcds so easily. E.g : $(2,6) = 2$ and $(1,4) = 1$ but $(3, 10) = 1$

krirkrirk
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  • Then it is very simple one, I hope. It is also the one I wanted avoid, if am correct. It is hopefully: $(a,a+b)=(b,a+b)=1$, taking product, get: $(ab, a+b) = 1$. I wanted to clear up on sum and product of $\gcd$, so wanted another approach. I am still not clear why sum is not valid, when product is. If some logical reasoning were provided kindly, then would serve the purpose a lot. – jiten Mar 07 '18 at 14:07
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    Right. Note that this is only simple because you asked for properties of the gcd. Now you might wonder why do those properties hold... – krirkrirk Mar 07 '18 at 14:09
  • Yes, that is the key to understanding why one operation(*) holds, but not the other(+), on $\gcd$. I would be waiting for your response on that, as it seems too crucial to understanding of $\gcd$. – jiten Mar 07 '18 at 14:10
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    I'd suggest you try a little bit on your own to prove those properties. Second one is easy, first one can be tough. See this question if you can't figure it out. – krirkrirk Mar 07 '18 at 14:17
  • Thanks for the pointer, but regarding the invalidity of sum operation on $\gcd$ request some help too. The link provided is for product operation only. – jiten Mar 07 '18 at 14:32
  • I'm not sure what help you're looking for ? It's simply not true, as shown in the counterexample in my answer. – krirkrirk Mar 07 '18 at 17:03