a relative prime triple 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?  
 A: 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 $\,\det(A) = -1$
A: 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 $c$. 
