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# 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

ANSWER: $|C|=0$. ## By Ashani Dasgupta

Ph.D. in Mathematics from University of Wisconsin Milwaukee (USA)
Founder - Faculty at Cheenta