<s>[INST] 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.
 [/INST] Let the roots of $x^2 - ax +b$ be $r$ and $s$, and the roots of $x^2 - bx + a$ be $u$ and $v$. Then
$$\begin{align}
r+s &= a\\
rs &= b\\
u+v &= b\\
uv &= a.
\end{align}$$
Let, without loss of generality, $a \leqslant b$. Thus
$$\begin{align}
uv &\leqslant u+v\\
\iff uv - u - v + 1 & \leqslant 1\\
\iff (u-1)(v-1) & \leqslant 1.
\end{align}$$
So $u = 1$ or $v = 1$ or $u = v = 2$. But the roots are supposed to be distinct, hence $u = 1$ or $v = 1$. Without loss of generality, $u = 1$.
Thus $a = v = r+s$, $b = v+1 = rs$, so
$$rs = r+s+1 \iff (r-1)(s-1) = 2,$$
and that leaves $r = 2$, $s = 3$ (or vice versa), so $a = 5, b = 6$.
</s>