parmenides51 wrote: Find the rational roots (if any) of the equation $abx^2 + (a^2 + b^2 )x +1 = 0 , \,\,\,\, (a, b \in Z).$ $a=b=0$ gives no solution $a\ne 0$ and $b=0$ gives solution $-\frac 1{a^2}$ $a=0$ and $b\ne 0$ gives solution $-\frac 1{b^2}$ $a,b\ne 0$ : $x=0$ is never a solution. So let $x=\frac pq$ with $p\in\mathbb Z_{>0}$, $q\in\mathbb Z\setminus\{0\}$ and $p,q$ coprime Equation is $abp^2+(a^2+b^2)pq+q^2=0$ and so $p|q^2$ and so $p=1$ Equation becomes $ab+(a^2+b^2)q+q^2=0$ So $(a^2+b^2)^2-4ab=c^2$ for some $c\ge 0$ If $ab<0$, this implies $-4ab\ge 2(a^2+b^2)+1$, which is $2(a+b)^2+1\le 0$, and no solution If $ab>0$, this implies $4ab\ge 2(a^2+b^2)-1$, which is $2(a-b)^2\le 1$ and so $a=b$ And $(a^2+b^2)^2-4ab=c^2$ becomes $4a^2(a^2-1)=c^2$ and so $a^2=1$ and so $q=-1$ And so rational roots can be : $\boxed{\text{S1 : }-\frac 1{n^2}\text{ with }(a,b)=(n,0)\text{ or }(0,n)}$ $\boxed{\text{S2 : }-1\text{ with }(a,b)=(1,1)\text{ or }(-1,-1)}$