A problem by Anish Roy

The above equation can be rearranged and written as $a - \frac{a}{b} + b - \frac{b}{a}= 0$ $\Rightarrow a(1- \frac{1}{b})+ b(1-\frac{1}{a}) = 0 .$
Now there are three cases,
Case 1 : first term $> 0$ and second term $< 0$ ,
Case 2 : first term $> 0$ and second term $< 0$ ,
Case 3 : first term $= 0$ and second term $= 0$
If we consider case $1$ then we get that $1- \frac{1}{a} < 0$ , which implies that $a < 1$ which is not possible since $a$ is a positive integer. Similarly case 2 also does not hold. So the only possibility is case 3 where both the terms are equal to zero. Thus equating $1- (\frac{1}{b}) = 0$ and $1-(\frac{1}{a}) = 0$ , we get $a = 1$ and $b = 1.$ Therefore the value of $a^2 + b^2 = 1+1 = \boxed{2}.$