# $\dfrac{pq}{r}+\dfrac{qr}{p}+\dfrac{pr}{q}\in\Bbb Z\,\Rightarrow\, \dfrac{pq}{r},\dfrac{qr}{p},\dfrac{pr}{q}\in\Bbb Z$

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.
• is " Rational Root Test implies that its rational roots are integers" true only when leading coefficient is $1$, and which is with our question ($x^3$)? – mathlover Jan 25 '17 at 12:31
• @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.\$ – Bill Dubuque Jan 25 '17 at 15:11
• @Num If $\,x^3+bx^2+cx+d\,$ has all integer coefficients then every rational root is an integer, by RRT.. – Bill Dubuque Jan 25 '17 at 16:02