Let $$X = \{-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5\}$$ and let $$S$$ be the set of all pairs $$(a,b) \in X \times X$$ such that $$x^2 + ax + b$$ and $$x^3 + bx + a$$ have at least a common real zero. How many elements are there in $$S$$?

If $$r$$ is a common zero, subtracting $$r(r^2+ar+b)$$ from $$r^3+br+a$$ gives $$a(1-r^2) = 0$$, so either $$a = 0$$ or $$r = \pm 1$$. When $$a = 0$$ a common zero exists exactly for $$b \le 0$$, giving 6 pairs; $$r = 1$$ needs $$a+b = -1$$ and $$r = -1$$ needs $$b = a-1$$, giving 10 pairs each. The only pair common to more than one family is $$(0, -1)$$, so by inclusion and exclusion the count is $$6+10+10-1-1-1+1 = 24$$.