Find all positive integers $a$, $b\neq1$ such that each of the equations:-
$$x^2-ax+b=0$$ and $$x^2-bx+a=0$$
has distinct positive integer roots.
I have taken x and y as roots for first equation and p and q for the second equation and got two equations from the sum of roots and product of roots but I am not able to progress further.Any idea will be appreciated
Intuitively, this is going to be difficult. If $p$ and $q$ are the zeroes of a quadratic polynomial $x^2-ax+b$, and both are positive and reasonably large, then $b=pq$ will generally be a lot bigger than $a=p+q$. So switching them is unlikely to give another polynomial with positive real roots. The following proof is an attempt to make this intuition precise.
Suppose without loss of generality that $a \geq b$, and let $p > q$ be the zeroes of $x^2-ax+b$ (note that $p$ and $q$ are distinct, so this is the only case that needs considering). Then $p+q=a$, $pq=b$.
If $q > 2$, then $b = pq > 2p > p+q = a$, contradicting our initial assumption. So we must have either $q=1$ or $q=2$.
If $q=1$, then $a=p+1$ and $b=p$. So we want to find $p$ such that $x^2-px+p+1$ has distinct integer zeroes: i.e., such that its discriminant $p^2-4(p+1)$ is a positive perfect square. Note that $$ p^2-4(p+1)=p^2-4p-4=(p-2)^2-8 $$ The only two perfect squares that differ by $8$ are $1$ and $9$, and so we must have $(p-2)^2=9$. Hence $p=5$, leading to $a=6$, $b=5$. And indeed, $$ x^2-6x+5\\ x^2-5x+6 $$ both have distinct positive integer zeroes.
If $q=2$, then $a=p+2$ and $b=2p$. But we also took $a \geq b$ and $p >q$. If $q=2$, then $a\geq b$ only when $p \leq 2$, and $p > q$ only when $p > 2$. So there can be no examples of this form. (If we did not require the roots to be distinct, then we could take $p=q=2$, yielding $a=b=4$. And in fact $x^2-4x+4$ has two (nondistinct) positive integer zeroes.)
