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Suppose we have three positive integers $x=a+c$, $y=b+c$ and $z=a+b+c$, where $x, y, z$ are relatively prime. I think i am right in saying that this implies that $a$ and $b$ must be coprime. My reasoning being that if $a$ and $c$ shared a factor and $b$ and $c$ shared a factor, then if those factors were the same the original triple wouldn't be coprime, and so $a$ and $b$ must have distinct compositions. Firstly is that reasoning correct? If it is then secondly, I'm not sure what (if anything) the condition says about $c$ in relation to $a$, $b$ and $a+b$. At one point i was thinking that if $c$ is not coprime to $a+b$, then it must contain a prime factor that is not contained within $a$ or $b$, and i think that is possibly all we can say about c.

Indeed can anything be concluded about $c$ in relation to the other two?

user25025
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2 Answers2

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Reducing your gcd a la Euclid, i.e. $\, (n,j,\color{#c00}k) = (n,j,\color{#c00}{\bar k})\ $ if $\,\color{#c00}{k\equiv \bar k}\pmod{\!n}\:$ yields

$\begin{align} (x,y,z)=({a\!+\!c},b\!+\!c,\color{#c00}{a\!+\!b\!+\!c}) &= (\color{#0a0}{a\!+\!c},b\!+\!c,\color{#c00}a)\ \ \ {\rm by}\ \ \color{#c00}{a\!+\!b\!+\!c\equiv a}\!\!\!\!\pmod{\!b\!+\!c}\\ &=\ \ \ \ \ \, (\color{#0a0}c,\color{#90f}{b\!+\!c},a)\ \ \ {\rm by}\ \ \color{#0a0}{a\!+\!c\equiv c}\ \ \,\pmod{\!a}\\ &= \ \ \ \ \ \, (c,\color{#90f}b,a)\ \ \ \ \ \ \ \ \:\!{\rm by}\ \ \color{#90f}{b\!+\!c\equiv b}\ \ \ \pmod{\!c} \end{align}$

So $\,(x,y,z) = 1\!\iff\! (a,b,c)=1\,$, which is not equivalent to $\,(a,b)=1$, e,g, $\,a,b,c = 2,2,1$

Or $\ d\mid a,b,c\Rightarrow\, d\mid x,y,z.\ $ $\,d\mid x,y,z\,\Rightarrow\, d\mid a,b,c\ $ by $\, a = z\!-\!y,\, b = z\!-\!x,\, c = x\!+\!y\!-\!z\,$

This is a $3$-dim version of this method: $\gcd(v)\mid\gcd(Av)\mid \det(A)\gcd(v)$ for linear $A,\,$ where here $\,\det(A) = -1$

Bill Dubuque
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  • Thanks Bill. What you're saying is that all three are coprime by application of Euclid's algorithm(?). I'm very new to this stuff (and very bad at it), i'll work through it slowly and come back with any questions. – user25025 Dec 10 '19 at 03:05
  • @user25025 We proved that $x,y,z$ have the same gcd as $a,b,c$. In particular $x,y,z$ have no common factor $\iff $ $a,b,c$ have no common factor. – Bill Dubuque Dec 10 '19 at 03:07
  • of course...i keep falling into the trap of not viewing them all together (but rather paired up) – user25025 Dec 10 '19 at 03:11
  • While i'm still working through the modular arithmetic, i think i see that my original inclination was on the right track. Can i not argue the following: consider any two of $a,b,c$ and assume that they have a common factor, then if the third also shares this factor $x, y ,z$ can not be coprime. Thus $a,b,c$ can not share a common factor and must be coprime. QED? Along the same lines i think i can also say given $(a,b,c)=1$ there can be no factor in a+b that can cancel with a factor of c, so a+b is coprime to c? (or is that last statement showing my intuition is off the rails?) – user25025 Dec 10 '19 at 05:43
  • @user25025 It is clear that a common factor of $a,b,c$ is also a common factor of $x,y,z$. The converse is the trickier direction. If you need help with the mod arithmetic then let me know specifically what part is not clear and I will elaborate. – Bill Dubuque Dec 10 '19 at 05:50
  • I think i see it all now. the first step is just finding the remainder of the second and third integer, then the second step looks for the gcd of that remainder against the first term, and similarly for the last step. Given that originally i think i was making the mistake of pairing them up, a natural question would be to consider pairwise coprime rather than just coprime. Everything here should still work except we obtain the conclusion that if x,y,z are pairwise coprime then so must be a,b,c? – user25025 Dec 10 '19 at 22:14
  • Yes, just like in the euclidean algorithm we can mod out one arg by another without changing the ideal because $(a, b\bmod a) = (a, b-qa) = (a,b)$. The pairwise version isn't correct - you already have a counterexample in a comment on the question. – Bill Dubuque Dec 10 '19 at 23:47
  • The dint on my desk where i keep banging my head just got a little bigger. Anyway i think i can see why i keep falling into that trap, and for the most part have a much better understanding of the problem. I have questions regarding the conditions for them being pairwise coprime but perhaps that is better in a separate post with more context. Thanks for your patient and kind instruction Bill. Much obliged. – user25025 Dec 11 '19 at 02:47
  • @user25025 Glad to see you've got it now. Don't sweat it much - we all have to climb the learning curve and everyone gets stuck at various points – Bill Dubuque Dec 11 '19 at 04:54
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No. $x,y$ needing to be coprime, states implicitly that $a-b$ shares no factor other than 1, with either sum. Parity of both $a,b$ are either both opposite $c$, or have $a,b$ opposite parity of each other. In the latter, $c$ and $z$ are opposite parity. In the former, $c$ and $z$ share parity. Modulo 3 states $a,b$ can't both be the additive inverse of $c$.