Croatian Mathematical Society Competitions

Let the roots of $x^2+px+q=0$ be $r$ and $s$, with $r\ne s$ and both integers.

By Vieta's formulas: $r+s=-p$ $rs=q$

Since $p$ and $q$ are primes, $-p$ is the sum of two distinct integers, and $q$ is their product.

Let $r$ and $s$ be distinct integers. Since $q$ is prime, $rs=q$ implies that one of $r$ or $s$ is $1$ and the other is $q$, or one is $-1$ and the other is $-q$.

Case 1: $r=1$, $s=q$ (with $q$ prime, $q\ne 1$) Then $r+s=1+q=-p⟹p=-(1+q)$ But $p$ must be prime, so $-(1+q)$ is prime. Since $q$ is prime $\ge 2$, $1+q\ge 3$, so $-(1+q)$ is negative and prime. The only negative primes are $-2,-3,-5,\dots$ So $1+q=2⟹q=1$ (not prime), $1+q=3⟹q=2$, $1+q=5⟹q=4$ (not prime), $1+q=7⟹q=6$ (not prime), etc. So only $q=2$ gives $p=-3$ (which is prime).

Check: $x^2+(-3)x+2=x^2-3x+2=(x-1)(x-2)$, roots $1$ and $2$ (distinct integers).

Case 2: $r=-1$, $s=-q$ (with $q$ prime, $q\ne 1$) Then $r+s=-1-q=-p⟹p=1+q$ $p$ must be prime, so $1+q$ is prime. Try $q=2$, $p=3$ (prime), $q=3$, $p=4$ (not prime), $q=5$, $p=6$ (not prime), $q=7$, $p=8$ (not prime), etc. So only $q=2$ gives $p=3$ (prime).

Check: $x^2+3x+2=(x+1)(x+2)$, roots $-1$ and $-2$ (distinct integers).

Case 3: $r=q$, $s=1$ (already considered in Case 1). Case 4: $r=-q$, $s=-1$ (already considered in Case 2).

Therefore, the only pairs $(p,q)$ of prime integers such that the solutions of $x^2+px+q=0$ are two distinct integers are: $$(p,q)=(3,2)\text{ and }(p,q)=(-3,2)$$