Let $p,q,r$ be integers such that the symmetric sum of fractions $\dfrac{pq}{r}+\dfrac{qr}{p}+\dfrac{pr}{q}$ is an integer.

Prove that each of the numbers: $\dfrac{pq}{r},\dfrac{qr}{p},\dfrac{pr}{q}$ is an integer.

How to do this? Someone posted but deleted soon.


Hint $ $ The hypothesis implies $\,\large {\big(x-\frac{pq}r\big)\big(x-\frac{qr}p\big)\big(x-\frac{pr}q\big)}\,$ has all integer coefficients, therefore the Rational Root Test implies that its rational roots are integers.

  • $\begingroup$ is " Rational Root Test implies that its rational roots are integers" true only when leading coefficient is $1$, and which is with our question ($x^3$)? $\endgroup$ – mathlover Jan 25 '17 at 12:31
  • $\begingroup$ @Ayush Yes, if the leading coefficient is $\,c\,$ then RRT implies that the denominator $\,d\,$ of a least-terms rational root must divide $c$. Only the monic case $\,c = \pm1.\,$ has a unique positive divisor $(= 1),\, $ since *all* positive divisors $d$ of $c$ can occur as a denominator, e.g. $\,(dx-1)(ex^{n-1}-1)\,$ has root $\,x = 1/d.\ $ $\endgroup$ – Bill Dubuque Jan 25 '17 at 15:11
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    $\begingroup$ @Num If $\,x^3+bx^2+cx+d\,$ has all integer coefficients then every rational root is an integer, by RRT.. $\endgroup$ – Bill Dubuque Jan 25 '17 at 16:02
  • $\begingroup$ @Num All the rational roots are integral (this is vacuously true.when it has no rational roots). $\endgroup$ – Bill Dubuque Jan 25 '17 at 16:07

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