Let $p,q,r$ be integers such that:$\dfrac{pq}{r}+\dfrac{qr}{p}+\dfrac{pr}{q}$ 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.
Proving a s sum of three symmetric fractions is an integer
2
$\begingroup$
algebra-precalculus
elementary-number-theory
1 Answers
5
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.
-
0is " Rational Root Test implies that its rational roots are integers" true only when leading coefficient is $1$, and which is with our question ($x^3$)? – 2017-01-25
-
0@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.\ $ – 2017-01-25
-
1@Num If $\,x^3+bx^2+cx+d\,$ has *all* integer coefficients then every rational root is an integer, by RRT.. – 2017-01-25
-
0@Num All the *rational* roots are integral (this is vacuously true.when it has no rational roots). – 2017-01-25