3
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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