Prove $a/b+b/a$ for $a$ and $b$ natural is only natural for $a=b$ Is it possible to prove that for any natural $a,b$ the value of $a/b+b/a$ will not be natural with exception $a=b$?
$$\begin{align}
{a\over b}+{b\over a}=k&amp;\implies a^2-kba+b^2=0\\
&amp;\implies a=\left(k\pm\sqrt{k^2-4}\over2\right)b\\
&amp;\implies\sqrt{k^2-4}\in\mathbb{N}\\
&amp;\implies k=2\\
&amp;\implies a=b
\end{align}$$
That's easy. $\frac{a}{b}+\frac{b}{a} = \frac{ a^2 + b^2 }{ab}$ So that $a/b+b/a$ is a natural number means $ a^2 +b^2 = k \cdot ab $ for some natural number k. We know the equation $x^2 -kx +1 =0$ has two integer roots if and only if $k=2$. To satisfy the condition, a should be equal to b.