<s>[INST] Show that if $a \neq b$ and a and b are positive then $\frac{a}{b}+\frac{b}{a}$ is never an integer 
Some observations I made is for $\frac{a}{b}+\frac{b}{a}$, is either:


*

*the denominator has to be one,

*the numerator has to be a multiple of the denominator or 

*the numerator and denominator have to be the same. 
Obviously, with the given conditions case 3 is eliminated since  $a \neq b$.
For case 1, if $b=1$, then $a=1$ which is a contradiction to the given condition that says $a \neq b$.
For case 2, I would think of examples. For example if $b=2$ then a multiple is $4$, so $a=4$. Then we have:
$$\frac{a}{b}+\frac{b}{a}=\frac{4}{2}+\frac{2}{4}=\frac{10}{4}$$
which is not an integer, but how would i proceed to show this case for all integers.
 [/INST] Presumably you want $a,b$ to be integers.  If $x = a/b$, solve the equation 
$$x + \dfrac{1}{x} = n$$
to get $$x = \dfrac{n}{2} \pm \dfrac{\sqrt{n^2-4}}{2} $$
The question is whether $\sqrt{n^2-4}/2$ is ever rational.  This would require $\sqrt{n^2-4}$ to be an integer (because the square root of an integer is only rational if it is an integer).  But then if $m = \sqrt{n^2-4}$, we would have $n^2 - m^2 = 4$.  The only squares of integers that differ by $4$ are $2^2$ and $0^2$, but that corresponds to $x = 1$, i.e. $a = b$.
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