Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 1 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.