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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 \(8\).

So, we get a contradiction.

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: \(|C|=0\).

 

 

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