Polynomial Problem (RMO 2015 Solution of Problem 2)

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.