Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $a,b,c>0$ be pairwise relatively prime and $n>2$ be odd. Can the equation, $a^n\cdot x^2+b^n\cdot x+c^n=0$, have rational roots $x$?

share|cite|improve this question
Start with finding those roots, with: $$x_{1,2}=\frac{-b^n\pm\sqrt{b^{2n}-4a^nc^n}}{2a^n}$$ – Salech Alhasov Mar 4 '12 at 18:55
Alternatively, I started to think of this problem as applying – user2468 Mar 4 '12 at 19:18

The roots are $$ \frac{-b^n + \sqrt{b^{2n}-4a^n c^n}}{2a^n}, \frac{-b^n - \sqrt{b^{2n}-4a^n c^n}}{2a^n} $$ Let's rule out the trivial case $b^{2n} = 4a^n c^n$. If $\sqrt{b^{2n}-4a^n c^n}$ is rational then so do the roots. The question boils down to showing the existence of $r \in \mathbb{Q}$ such that $r^2 = b^{2n}-4a^n c^n$ under the conditions your gave for $a,b,c,n$. Not sure yet how to answer this.

share|cite|improve this answer
So we have $b^{2n}-r^2=4(ac)^n$ or $(b^n-r)(b^n+r)=4(ac)^n$, where $r$ is rational. Is this possible? – Craig Feinstein Mar 4 '12 at 19:54
@CraigFeinstein Exactly. But I can't go anywhere from there. – user2468 Mar 4 '12 at 20:08
@CraigFeinstein At least this implies that $r$ is an integer. For suppose $r=p/q$ is a reduced fraction and a prime $m$ divides $q$ then $(q b^n-p)(q b^n+p) = 4q^2(ac)^n$ and $m$ divides the RHS but not the LHS. – WimC Mar 4 '12 at 20:48

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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