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.

Problem:

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

Discussion:

$ P(-1) = 1-a+b , P(0) = b, P(1) = 1 + a + b $ and $ s=2 $.
According to the problem

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

Clearly $ \dfrac {(1-a+b)^2 + (1+a+b)^2}{2} = b^2 $ and $ s=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  $ and $ s=2$
$ \Rightarrow 1 + a^2 +2b = 0 $ and $ s=2$
$ \Rightarrow 2b = -(1+a^2) $ and $ s=2$ implying b is negative.

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

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

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

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

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

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

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

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

Right hand side is rational. Hence left hand side is also rational. This implies $latex \sqrt c $ and $ s=2 $ is rational. Since c is an integer, this implies $ \sqrt c $ and $ s=2 $ is integer. Hence b is integer.

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

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

Chatuspathi:

• Paper: RMO 2015 (West Bengal)
• What is this topic: Polynomial
• What are some of the associated concepts: Rational and surd
• Where can learn these topics: Cheenta I.S.I. & C.M.I. course,Cheenta Math Olympiad Program, discuss these topics in the ‘Polynimial’ module.
• Book Suggestions: Polynomials by Barbeau