Number of $(p,q)$ in $\mathbb R^2$ such that $x^4+px^2+q$ is divisible by $x^2+px+q$ I have found four through my attempt, but apparently the answer is that there are five pairs.
Mine were $(-1,0),(-1,1),(0,0),(0,1)$.
What am I missing?

Fun problem here's another approach using Vieta's formulas:
The sum of the roots of $x^2+px+q$ is $-p$.  The sum of the roots of $x^4+px^2 + q$ is $0$. Similarly the product of the roots of both polynomials is $q$.  In particular this tells us that if $x^2+px+q$ divides $x^4+px^2 + q$ the quotient would have to be $x^2-px+1$, unless $q=0$.  
If $q=0$ the roots of $x^2+px+q$ are $0$ and $-p$. Zero is obiously a root of $x^4+px^2$, so we just need to plug $-p$ in and see when it's also a root. This gives us the equation $p^4+p^3 =0$. So we get two solutions $(p,q)= (0,0)$ and $(p,q)= (-1,0)$.
Suppose $q \ne 0$ now we can look at the coefficients of $x^2$ and $x$ in $(x^2+px+q)(x^2-px+1)$, and set them equal to $p$ and $0$ respectively.  This gives us the equations:
$$q-p^2+1 = p$$
$$p-qp = 0$$
The second equation forces either $q=1$ or $p =0$, which if we plus into the other equation and solve gives us the ordered pairs $(p,q) = (-2,1)$, $(1,1)$, and $(0,-1)$.
