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December 7, 2015

West Bengal RMO 2015 Problem 2 Solution - Polynomial Problem

The second stage examination of INMO, the Regional Mathematical Olympiad (RMO) is a three hour examination with six problems. The problems under each topic involve high level of difficulty and sophistication. The book, Challenge and Thrill of Pre-College Mathematics is very useful for preparation of RMO. West Bengal RMO 2015 Problem 2 Solution has been written for RMO preparation series.


Let P(x)=x^2+ax+b be a quadratic polynomial where a,b are real numbers. Suppose l\angle P(-1)^2,P(0)^2,P(1)^2r\angle  be an arithmetic progression of positive integers. Prove that a,b are integers.


P(-1) = 1-a+b , P(0) = b, P(1) = 1 + a + b  .
According to the problem

(1-a+b)^2 , b^2 , (1+a+b)^2 are in arithmetic progression of positive integers.

Clearly \dfrac {(1-a+b)^2 + (1+a+b)^2}{2} = b^2

This implies \dfrac {1 + a^2 + b^2 -2a +2b -2ab + 1+ a^2 + b^2 +2a +2b + 2ab}{2} = b^2
\Rightarrow 1 + a^2 + b^2 +2b = b^2
\Rightarrow 1 + a^2 +2b = 0
\Rightarrow 2b = -(1+a^2) implying b is negative.

Now we know (1-a+b)^2  is an integer.

Then 1+a^2 +b^2 -2a +2b -2ab = -2b + b^2 -2a +2b -2ab  (replacing 1+a^2 = -2b  )

This implies b^2 -2a -2ab  is an integer. But b^2 is also an integer. Hence 2a + 2ab  is an integer.

Now we also know (1+a+b)^2  is an integer.

Then 1+a^2 +b^2 +2a +2b +2ab = -2b + b^2 + 2a +2b + 2ab  (replacing 1+a^2 = -2b  )

Again replacing -(1+a^2) = 2b  we get b^2 + 2a - a(1+a^2)  is an integer or (a - a^3)  is an integer.

Note that b^2  is some positive integer. Let it be b^2 = c  . Then b= - sqrt c  where c is some positive integer (as we know b is negative)

1+a^2 = 2\sqrt c  or a^2 = 2 \sqrt c - 1
a(1-a^2) = k  (suppose). Then a(1- (2 \sqrt c - 1)) = k  or 2a(1-\sqrt c) = k
squaring both sides we get
4a^2 (1+c - 2\sqrt c) = k^2
\Rightarrow 4(2 \sqrt c - 1) (1+ c - 2 \sqrt c) = k^2
\Rightarrow 4(2 \sqrt c + 2 c \sqrt c - 4c - 1 - c + 2 \sqrt c) = k^2
\Rightarrow 4(4 \sqrt c + 2c \sqrt c - 5c - 1) = k^2
\Rightarrow (4+2c)\sqrt c = \dfrac{k^2}{4} + 5c + 1
\Rightarrow \sqrt c = \dfrac{k^2 + 20c + 4}{4(4+2c)}

Right hand side is rational. Hence left hand side is also rational. This implies \sqrt c  is rational. Since c is an integer, this implies \sqrt c  is integer. Hence b is integer.

We know a^2 = -2b - 1  . Since b is integer, therefore a^2  is integer.
Again a(1-a^2)  is integer and a^2  is integer, implies a must be rational.

Finally, if a is rational and a^2 is integer then a must be integer.


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