1. Let the quadratic be $p(x)=x^2+ax+b$ with roots $r_1,r_2$ (possibly equal).

   The condition says:

   whenever $\alpha$ is a root, $\alpha^2-2$ is also a root.

   So the set of roots must be closed under the map $f(x)=x^2-2.$

2. Since a quadratic has at most two roots, starting from any root $\alpha$, the numbers $\alpha,\ f(\alpha),\ f(f(\alpha)),\dots$ must remain among at most two values.

   Thus the root set must consist of fixed points or a 2-cycle of the map $f(x)=x^2-2$.

Step 1: Fixed points

We solve $x^2-2=x$ which gives $x^2-x-2=0$ $(x-2)(x+1)=0.$ So fixed points are $x=2,-1.$

This gives possible root sets:

• $\{2\}$
• $\{-1\}$
• $\{2,-1\}$

These correspond to the quadratics:

• double root $2$: $(x-2)^2=x^2-4x+4$
• double root $-1$: $(x+1)^2=x^2+2x+1$
• roots $2,-1$: $(x-2)(x+1)=x^2-x-2$

So far, $3$ pairs $(a,b)$.

Step 2: 2-cycles

Now solve $f(f(x))=x,$ with $f(x)\ne x$.

Since $f(x)=x^2-2,$ we get $f(f(x))=(x^2-2)^2-2=x.$ So $(x^2-2)^2-2=x$ $x^4-4x^2-x+2=0.$

Factor this polynomial. Since fixed points $x=2,-1$ must be roots, divide by $(x-2)(x+1)$: $x^4-4x^2-x+2=(x-2)(x+1)(x^2+x-1).$

Thus the other solutions are $x^2+x-1=0,$ so $x=\frac{-1\pm\sqrt5}{2}.$

Let $r_1=\frac{-1+\sqrt5}{2},\qquad r_2=\frac{-1-\sqrt5}{2}.$ These are distinct and satisfy $f(r_1)=r_2,\qquad f(r_2)=r_1,$ so they form a 2-cycle.

Hence another valid quadratic is $(x-r_1)(x-r_2)=x^2-(r_1+r_2)x+r_1r_2.$ Now