AustriaMO2011

Without loss of generality, we assume $x_1=x_2+1$. By Vieta's formulas, this implies $p=-(x_1+x_2)=-2x_2-1$ and $q=x_1{x}_2=x_2^2+x_2$. Therefore, it is enough to check that $x_2$ has to be an integer.

Case 1: $q=p-1$ This implies $x_2^2+x_2=-2x_2-1-1$ and therefore $x_2^2+3x_2+2=0$ and $x_2=-1$ or $x_2=-2$, both of which are integers.

Case 2: $q=p+1$ We find $x_2^2+x_2=-2x_2-1+1$ and therefore $x_2^2+3x_2=0$ and $x_2=0$ or $x_2=-3$, both of which are integers.