frac{a}{b}+\frac{b}{a} \notin \mathbb{Z}$

Question

I know that $a \neq b$, then $\frac{a}{b}+\frac{b}{a}$ would not equal to $m$ (an integer) so I set them into one that wasn't a fraction by squaring. So I got $\frac{a^2+b^2}{ab}$. How can I show that $a^2+b^2$ is not divisible by $ab$?

Answer 1

I assume $a,b\in \Bbb Q-\{0\}$ unless there are infinitely many counterexamples.

Let $c={a\over b}. $If $${a\over b}+{b\over a}=c+{1\over c}=m$$then $$c^2-mc+1=0$$therefore $$c={m\pm\sqrt{m^2-4}\over2}$$which is quotient number only if $m^2-4$ is a perfect square, which is impossible since $$(m-1)^2<m^2-4<m^2$$for $m>2$. For $m=0,1$, $c\notin \Bbb R$ and for $m=2$, $a=b$ which is a contradiction.