Find $(a,b)$ if $x^2-bx+a = 0, x^2-ax+b = 0$ both have distinct positive integers roots 
If $x^2-bx+a = 0$ and $x^2-ax+b = 0$ both have distinct positive integers roots, then what is $(a,b)$?

My Try:
$$\displaystyle x^2-ax+b = 0\Rightarrow x = \frac{a\pm \sqrt{a^2-4b}}{2}$$
So here $a^2-4b$ is a perfect square.
Similarly 
$$\displaystyle x^2-bx+a = 0\Rightarrow x = \frac{b\pm \sqrt{b^2-4a}}{2}$$
So here $b^2-4a$ is a perfect square.
But I did not understand how can I solve after that.

Would a=b=4 work? In that case the (double) solution in both cases is 2 and that's positive
