# elementary divisibility argument

I am trying to argue that for distinct primes $p,q,r$ we have that

$$\gcd (pq + qr+ pr ,pqr ) = 1 = g$$

and I am wondering whether people find the following argument convincing :

Consider the prime decompositions; for $g$ to be greater than $1$ we need at least one of $p,q,r$ to occur in the decomposition of $pq + qr+ pr$. Without loss of generality suppose it is $p$ ( not necessarily exclusively ). But then the remainder modulo $p$ must be zero. However the remainder is $qr$ which is not zero modulo $p$ as $p,q,r$ are all mutually coprime. ( using $x\mid (a+b)$ and $x\mid a$ implies $x\mid b$ ) So the result follows.

-
It is probably a matter of taste, but once you get to the point that $p$ divides $pq+pr+qr$, you might note that $p$ divides $pq$ and $pr$, and thus divides $qr$. Since $p$ is prime, it must divide $q$ or $r$, and thus since $q$ and $r$ are primes, it must equal one of them, contradicting the assumptions. –  Andreas Caranti Jan 31 '13 at 11:43
That seems legit. –  Tunococ Jan 31 '13 at 11:43

Yes, that works. In fact it's true iff $\rm\:p,q,r\:$ are pair-coprime. Indeed, if all pairs have gcd $= 1$ then $\rm\:(pq\!+\!qr\!+\!rp,p) = (pq\!+\!qr\!+\!rp\ mod\ p,p)= (qr,p) = 1\:$ since, by Euclid's lemma, $\rm\,(q,p)=1=(r,p)$ $\,\Rightarrow\,$ $\rm (qr,p)=1.\,$ By symmetry $\rm\:pq\!+\!qr\!+\!rp\,$ is also coprime to $\rm\,q\,$ and $\rm\,r,\,$ thus it is coprime to the product $\rm\:pqr,\,$ again by Euclid. Conversely, if some pair has gcd $\rm\,d> 1\,$ then it is clear that $\rm\,d\,$ divides the given gcd: e.g. $\rm\:d\,|\,\bf\color{#0A0}q,\color{#C00}r\,$ $\Rightarrow$ $\rm\,d\,|\,pq{\bf\color{#C00}r}, (p\!+\!r){\bf\color{#0A0}q}\!+\!{\bf\color{#C00}r}p,\,$ so $\rm\,d\,$ divides their gcd.