Pre RMO 2015

From the above equation

$a- \dfrac{a}{b}=\dfrac{b}{a}-b$

$a (1- \dfrac{1}{b})=b (\dfrac{1}{a}-1)$

When $a,b$ are positive integers and $\dfrac{1}{b}$ is less than $1$

So, $(\dfrac{1}{a}-1)>=0$

But it can.t be greater than zero so $a=1$ and $b=1$

Apply reasoning,

Since LHS is a positive integer RHS must be so.

So both $\displaystyle \frac{a}{b},\frac{b}{a}$ are integers. Hence, $a|b$ & $b|a$ which guarantees $a=b$

So we have , $2a=2$ therefore $\boxed{a^2+b^2=2}$

$a+b=\frac{a}{b}+\frac{b}{a}$ $a+b=\frac{a^2+b^2}{ab}$ $a^2b+ab^2=a^2+b^2.......[1]$ $a^2b-a^2=b^2-ab^2$ $a^2(b-1)=b^2(1-a)$ $(\frac{a}{b})^2=\frac{1-a}{b-1}$ $\Rightarrow \frac{1-a}{b-1}\geq0$ $1-a\geq0$ $1\geq a$ As $a$ is postive integer $\therefore a=1$ Put this in $[1]$ $b\cancel{+b^2}=1\cancel{+b^2}$ $b=1$ $\Rightarrow a^2+b^2=1^2+1^2=\boxed{2}$ Note:

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