Post Number:#2 by tanmoy » Thu Jan 28, 2016 1:58 pm
Let $a$ and $b$ be the two distinct positive integral solutions,where $a<b$.Then $x^{2}-px+q=(x-a)(x-b)$ implying that $a+b=p$ and $ab=q$.But $q$ is prime,so,$a=1$.Therefore,$q=b$ and $p=b+1$.The only two consecutive primes are $2$ and $3$.So,$(p,q)=(3,2)$.