Question:

Let $$C \subset \mathbb{ZxZ}$$ be the set of integer pairs $$(a,b)$$ for which the three complex roots $$r_1,r_2,r_3$$ of the polynomial $$p(x)=x^3-2x^2+ax-b$$ satisfy $$r_1^3+r_2^3+r_3^3=0$$. Then the cardinality of $$C$$ is

A. $$\infty$$

B. 0

C. 1

D. $$1<|C|<\infty$$

Discussion:

We have $$r_1+r_2+r_3=-(-2)=2$$

$$r_1r_2+r_2r_3+r_3r_1=a$$ and

$$r_1r_2r_3=-(-b)=b$$

Also, of the top of our head, we can think of one identity involving the quantities $$r_1^3+r_2^3+r_3^3=0$$ and the three mentioned just above.

Let’s apply that:

$$r_1^3+r_2^3+r_3^3-3r_1r_2r_3=(r_1+r_2+r_3)(r_1^2+r_2^2+r_3^2-(r_1r_2+r_2r_3+r_3r_1))$$ …$$…(1)$$

Also, note that $$r_1^2+r_2^2+r_3^2=(r_1+r_2+r_3)^2-2(r_1r_2+r_2r_3+r_3r_1)$$

So $$r_1^2+r_2^2+r_3^2=(2)^2-2(a)=4-2a$$.

So by equation $$(1)$$,

$$0-3b=2(4-2a-a)$$

or, $$6a-3b=8$$.

Now, we notice the strangest thing about the above equation: $$3|(6a-3b)$$ but $$3$$ does not divide $$8$$.

Why did this contradiction occur? We didn’t start off by saying “assume that … something …”. Well, even though we pretend to not assume anything, we did assume that this equation has a solution for some $$a,b$$. So, the
ANSWER: $$|C|=0$$.