How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?

# Theory of Equation (TIFR 2014 problem 10)

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