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If we have $\left( a,b,c \right)=1$ then am i right in saying $\frac{a+b+c}{c}\notin \mathbb{N}$ and $c$ is coprime to $a+b+c$? My reasoning being that there is no common factor that can be taken out of the sum to cancel with a factor in the denominator. If $c$ wasn't in the numerator it would be a different story.

(the reason i ask is that my intuition using prime factors when dealing with these sort of problems in the past has lead me to incorrect conclusions - is it incorrect here?).

If the above is correct then based on similar arguments wouldn't $\frac{ab\left( a+b+c \right)}{c}\in \mathbb{N}$ imply $ab=dc$? This seems intuitive to me since i can see that $a,b$ could possibly share different factors of $c$ in such a way that their multiplication forms a multiple of $c$.

user25025
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  • To debug your reasoning we need to know precisely how you made the incorrect inference that $,(a,b,c)=1,\Rightarrow, 1 = (a+b+c,,c) = (a+b,c).\ $ Please elaborate. – Bill Dubuque Dec 16 '19 at 01:42
  • @BillDubuque Firstly thank you for linking to the other duplicate. Secondly i think i can see where it is i am going wrong. As you say I'm assuming that $\left( a,b,c \right)=1\Rightarrow \left( a+b,c \right)=1$, and P Vanchinathan example nicely demonstrates why that can't work. I can see that now. The reason I’m going wrong, is that i'm incorrectly assuming if $a,b,c$ share no common factors, there isn't a factor we can take out the front of $a+b+c$ to cancel with a factor in $c$. This neglects the fact that even if $(a,b)=1$ then $a+b$ still might have a common factor with $c$. – user25025 Dec 16 '19 at 23:22

1 Answers1

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Take $a,b$ to be coprime numbers then for any $c$ we have $\gcd(a,b,c)=1$. Now take $c>1$ to be a factor of $a+b$ (if prime, take $c$ as $a+b$). Now $(a+b+c)/c$ will be an integer. Example: $a=5,b=7, c=4$. There are examples not of this kind too: $a=4, b=8, c= 3$.

P Vanchinathan
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  • Thanks for the reply. I'm not sure where my reasoning is falling down here. Consider https://math.stackexchange.com/questions/388560/is-the-sum-of-two-coprime-numbers-also-coprime-to-its-summands why does the addition of a third number change the reasoning? – user25025 Dec 16 '19 at 00:55
  • @user25025 That question is a dupe - now linked to many more answers which you may find helpful. – Bill Dubuque Dec 16 '19 at 01:08
  • @BillDubuque Hi Bill. Not sure i follow. are you saying my question is a duplicate of the one i linked in my comment? i'm guessing it probably is, but i just don't see why an argument along the lines of not being able to pull out a common factor across all three isn't enough to imply the sum isn't coprime to a summand (as P Vanchinathan's example demonstrates). I guess another way would be to explain what condition i need for $a+b+c$ to be coprime to $c$. Or am i making little sense? – user25025 Dec 16 '19 at 01:21
  • @user25025 No, the dupe link is for the other question. I mentioned it here since you may find the newly linked answers instructive. – Bill Dubuque Dec 16 '19 at 01:43