<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] If $a,b$ are not integers, nothing stops $x+\frac1x$ being an integer. In fact, that happens infinitely many times as the function is continuous.

Let $d=gcd(a,b)$ and $a=xd, b=yd$ with $gcd(x,y)=1$. Then
$$\frac{a}{b}+\frac{b}{a}=\frac{x^2+y^2}{xy}.$$
We will show that if $p$ is a common prime divisor of $x^2+y^2$ and $xy$ then $p=2$. Indeed, 
$$(x+y)^2=x^2+y^2-2xy\text{ and }(x-y)^2=x^2+y^2-2xy$$
are both multiples of $p$. Thus $p$ divides both $x+y$ and $x-y$, and so $p$ divides $2x$ and $2y$. Since $gcd(x,y)=1$, $p$ divides $2$ and $p$ must be $2$.
It follows that $x^2+y^2$ and $xy$ are both even. From $xy$ is even, we see that either $x$ or $y$ is even. That together with $x^2+y^2$ is even, we see that both $x$ and $y$ are even. It contradicts the fact that $gcd(x,y)=1$.
Thus $x^2+y^2$ and $xy$ don't have a common prime factor. $\dfrac{x^2+y^2}{xy}$ is an integer only if $xy=1$, equivalently $x=y=\pm1$, or $a=b$.
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