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 $ab$?
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https://math.stackexchange.com/questions/1468189/show-that-if-a-neq-b-and-a-and-b-are-positive-then-fracab-fracba, https://math.stackexchange.com/questions/1752529/a-b-b-a-is-an-integer-if-and-only-if-a-b – Martin R Jun 12 '20 at 19:19
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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.
Mostafa Ayaz
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I am confused on the bottom part. I undersatnd that it can only be an integer if m^2-4 is a perfect square, but i dont understand anything after that – Jun 12 '20 at 20:52
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The implication is that a perfect square cannot exist between two consecutive perfect squares. – Mostafa Ayaz Jun 13 '20 at 05:43