Find $(a^2+b^2),$ where $(a+b)=\dfrac{a}{b}+\dfrac{b}{a}$ The question is the same .Find $a^2+b^2$. I think we have to find $a$ and $b$ firstly. Given that $a$ and $b$ are integers.

$$a+b=\frac{a}{b} + \frac{b}{a}$$
therefore,$$a^2+b^2=ab(a+b)$$
or,$$a^2(1-b)+b^2(1-a)=0$$
from this relation we can say $1-b=1-a=0$
therefore,$$a=b=1$$
so finally,
$$a^2+b^2=2$$
