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# 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 $P(x)=x^2+ax+b$ 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.

## Discussion: $P(-1) = 1-a+b , P(0) = b, P(1) = 1 + a + b$.
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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