Find all prime numbers $p, q, r$ such that \[15p+7pq+qr=pqr.\]
Problem
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Tags: number theory proposed, number theory
mathmdmb
14.11.2010 15:03
Thank you a lot for the problem,it seems t be nice.Will you give a link to all problems of Slovenia-2010.
Amir Hossein
14.11.2010 15:10
mathmdmb wrote: Thank you a lot for the problem,it seems t be nice.Will you give a link to all problems of Slovenia-2010. You're welcome. I haven't posted all of the problems. I'll give you the link when I'm done
Martin N.
14.11.2010 15:13
amparvardi wrote: Find all prime numbers $p, q, r$ such that \[15p+7pq+qr=pqr.\] Rewriting the equation as $p(qr-7q-15)=qr$, we see that either $p=q$ or $p=r$. In the first case, we get $pr-7p-15=r\Leftrightarrow (p-1)(r-7)=22$, giving the solution $(2,2,29)$ in primes. The second case yields $pq-7q-15=q\Leftrightarrow 15=q(p-8)$ and the other two solutions $(13, 3, 13)$ and $(11,5,11)$.
Amir Hossein
14.11.2010 15:16
Martin N. wrote: ...and the only solution in primes $(13, 3, 13)$. Hmmm... What about $(2,2,29)$ and $(11, 5, 11)$?
WakeUp
14.11.2010 15:20
Ummm.....
$p|qr$ so $p=q$ or $p=r$.
We first suppose $p=q$. Then $15p+7p^2+pr=p^2r$ or $15+7p+r=pr$, rearrange to $(p-1)(r-7)=22=2\cdot 11$.
$p-1>0$ so both factors are positive and in particular $r>7$. If $p,r$ are both odd then the $LHS$ is divisible by $4$, a contradiction. But $r>7$ meaning $r\not= 2$ so that $p=2$ and we obtain the solution $(2,2,29)$.
Now suppose $p=r$ so that the equation becomes $15p+7pq+pq=p^2q\implies 15+8q=pq$. Hence $q|15$ so $q=3$ or $q=5$.
Substituting these back into the equation gives the solutions $(13,3,13)$ and $(11,5,11)$.
So all solutions are $(2,2,29)$, $(13,3,13)$ and $(11,5,11)$.
Martin N.
14.11.2010 15:44
amparvardi wrote: Martin N. wrote: ...and the only solution in primes $(13, 3, 13)$. Hmmm...What about $(2,2,29)$ and $(11, 5, 11)$? Uups sorry...$15+7=22$ and not $23$^^ I'll edit it, thank you.
rafaello
27.11.2020 22:46
We have that $15p+7pq+qr=pqr.$ Thus, we have that $p\mid qr\implies p=q \text{ or } p=r$. If $p=q$, then $15+7p+r=pr\implies (p-1)(r-7)=22=2\cdot 11$ Thus, $p=2,3,12, 23$, which means that corresponding values for $r$ are $29,18,9,8$, we see that only solutions from here is $(p,q,r)=(2,2,29)$. If $p=r$, then $15+8q=pq\implies q=3 \text{ or } q=5$, hence corresponding values for $p$ are $13$ and $11$, both are solutions hence we have $(p,q,r)=(13,3,13)$ and $(p,q,r)=(11,5,11)$.