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?
 A: One way of doing the polynomial division is to use an upgraded form of the remainder theorem. Dividing by $x^2+px+q$ is equivalent to setting $x^2=-px-q$. Here that works nicely because the original polynomial is a polynomial in $x^2$.
So the remainder is computed first as $$(-px-q)^2-p(px+q)+q=p^2x^2+(2pq-p^2)x+(q^2-pq+q)$$Substituting for $x^2$ again gives $$p^2(-px-q)+(2pq-p^2)x+(q^2-pq+q)=(-p^3+2pq-p^2)x+(-p^2q+q^2-pq+q)$$
Now the remainder is required to be zero so the coefficient of $x$ is zero $$-p^3+2pq-p^2=-p(p^2+p-2q)=0$$And this gives $p=0$ or $p^2+p-2q=0$
The constant term is also zero, so $$−p^2q+q^2−pq+q=-q(p^2+p-q-1)=0$$
so $q=0$ or $p^2+p-q-1=0$
So then it is a question of testing the cases. The only difficulty is with the two quadratic factors simultaneously - the pairs with $p$ or $q$ or both equal to zero easily give three solutions. Subtracting one quadratic from the other gives $q=1$ and then there are two solutions for $p$.

For setting $x^2=-px-q$ note that polynomial division will give us $$x^4+px^2+q=(x^2+px+q)s(x)+r(x)$$
for some polynomials $s(x), r(x)$. Here the degree of $r(x)$ as the remainder is less than the degree of $x^2+px+q$, so $r(x)$ has no quadratic or higher terms. I am not interested at all in $s(x)$. If I put $x^2=-px-q$, the term involving $s(x)$ vanishes and I'm left with $r(x)$, which is unchanged.
A: Hint $ $ If $\,f(x)= x^2\!+px\!+\!q\,$ has roots $\,x=a,b\,$ then$\,f(x^2) = f(x)g(x)$ $\Rightarrow\,f(a^2) = 0 = f(b^2).\,$ Thus $\,a,b,a^2,b^2\,$ are all roots of $f$. But a quadratic has at most two roots so we must have $\,a^2 = a\,$ or $\,a^2 = b,\,$ etc. Solve for $\,a,b\,$ in all cases (easy!) then use $\,(x\!-\!a)(x\!-\!b) = x^2\!+px+q\,$ to get all $\,p,q.$
A: Polynomial division of your two polynomials gives reminder $q-pq-p^2q+q^2+(-p^2-p^3+2pq)x$. You need both terms, on $x^{0}$ and $x^{1}$ zero. Solving this system gives me $(-2,1)$, $(-1,0)$, $(0,-1)$, $(0,0)$, $(1,1)$.
A: 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 $(0,-1)$.
