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.

(I assume that $a$ and $b$ are both integers...)
Let $x=\dfrac{a}{b}+\dfrac{b}{a}=\dfrac{a^2+b^2}{ab}$. We can assume that $a$ and $b$ are relatively prime, since substituting $a/g$ and $b/g$ instead of $a$ and $b$ doesn't change the value of $x$.
If $x$ is integer, than $ab$ should divide $a^2+b^2$, so
$$a|(a^2+b^2) \; \Rightarrow\; a|b^2$$
$$b|(a^2+b^2) \; \Rightarrow\; b|a^2$$
Since we assumed that $a$ and $b$ are relatively prime, these formulas hold only when $a=b=1$, which is contradicted to the condition $a\neq b$.
Therefore, $x\not\in \mathbb{Z}$.
