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.

Suppose $x^2-ax+b$ and $x^2-bx+a$ have positive integer roots. Then
$$x^2-ax+b=(x-c)(x-d)\qquad\text{and}\qquad x^2-bx+a=(x-e)(x-f),$$
for some positive integers $c$, $d$, $e$ and $f$. As the roots are must be distinct, we may assume without loss of generality that $c>d$ and $e>f$. Now compare coefficients.
Hint 1:

 It follows that $c+d=a=ef$ and $cd=b=e+f$.

Hint 2:

 If $c>d>1$ then $cd>c+d$.

Hint 3:

 If $d>1$ and $f>1$ then $cd>c+d=ef>e+f=cd$, a contradiction.

Hint 4:

 Without loss of generality we have $f=1$, so $e=c+d$ and $e+1=cd$.

Hint 5:

 It follows that $(c-1)(d-1)=2$, so $c=3$ and $d=2$, and hence $e=5$.

